thought stringlengths 140 6.42k | answer stringlengths 1 44 |
|---|---|
The expression inside the ceiling function is 3(6β 2 1 β ). First, calculate the value inside the parentheses: 6β 2 1 β =5.5. Next, multiply by 3: 3Γ5.5=16.5. The ceiling function βxβ is defined as the smallest integer greater than or equal to x. For x=16.5, the smallest integer nβ₯16.5 is 17. | 17 |
To find the value of the expression 6 x 4 +2y 2 β given x=2 and y=5, we substitute the values into the formula. First, calculate x 4 : 2 4 =2Γ2Γ2Γ2=16. Next, calculate 2y 2 : 2Γ(5 2 )=2Γ25=50. Now, add these results together: 16+50=66. Finally, divide by 6: 6 66 β =11. | 11 |
The cake costs 6 euros. Emily has 5 USD. Given the exchange rate is 1 euro = 1.25 USD, we first convert Emily's 5 USD into euros. To do this, we divide the amount in USD by the exchange rate: 5/1.25=4. So, Emily has 4 euros. To find out how much Berengere needs to contribute, we subtract Emily's contribution from the t... | 2 |
First, simplify the expression inside the first cube root: 1+8=9. So, the first part is 3 9 β . Next, simplify the expression inside the second cube root: 3 8 β =2, so 1+ 3 8 β =1+2=3. The second part is 3 3 β . Now, multiply the two parts together: 3 9 β β
3 3 β = 3 9β
3 β = 3 27 β . Since 3 3 =27, the cube roo... | 3 |
To find g(f(β2)), we first calculate f(β2) by substituting β2 into the function f(x)=x 3 +3. This gives f(β2)=(β2) 3 +3=β8+3=β5. Next, we substitute this result into the function g(x)=2x 2 +2x+1, so we need to find g(β5). Calculating this yields g(β5)=2(β5) 2 +2(β5)+1=2(25)β10+1=50β10+1=41. | 41 |
For the function f(x) to be continuous at x=2, the limit of f(x) as x approaches 2 from the left must equal the limit as x approaches 2 from the right. The value from the left (and at x=2) is 2(2) 2 β3=2(4)β3=8β3=5. The value from the right is a(2)+4=2a+4. Setting these equal gives 2a+4=5. Subtracting 4 from both sides... | 1/2 |
For the function f(x) to be continuous at x=3, the limit as x approaches 3 from the left must equal the limit as x approaches 3 from the right, which must also equal f(3). From the left and at x=3: f(3)=3(3) 2 +2=3(9)+2=27+2=29. From the right: lim xβ3 + β f(x)=a(3)β1=3aβ1. For continuity, we set these equal: 3aβ1=29.... | 10 |
First, find the rate of one faucet. Three faucets fill 100 gallons in 6 minutes, so 1 faucet fills 100/3 gallons in 6 minutes. The rate of one faucet is R=(100/3)/6=100/18=50/9 gallons per minute. Now, we have six faucets, so their combined rate is 6Γ(50/9)=300/9=100/3 gallons per minute. We need to fill a 25-gallon tu... | 45 |
To find the point where the line intersects the y-axis, we first need the equation of the line. The slope m is calculated as m= x 2 β βx 1 β y 2 β βy 1 β β = 3β1 11β7 β = 2 4 β =2. Using the point-slope form yβy 1 β =m(xβx 1 β ) with the point (1,7), we get yβ7=2(xβ1). Simplifying this to slope-intercept form ... | (0, 5) |
To find the coefficient of the x 2 term in the expansion of (ax 3 +3x 2 β2x)(bx 2 β7xβ4), we identify all pairs of terms (one from the first polynomial and one from the second) whose product results in x 2 . Let the first polynomial be P(x)=ax 3 +3x 2 β2x and the second be Q(x)=bx 2 β7xβ4. The terms in P(x) are ax 3 , ... | 2 |
We are given f(3)=1 and the functional equation f(2x)=2f(x). We want to find f β1 (64), which is the value x such that f(x)=64. Starting from f(3)=1, we can apply the rule f(2x)=2f(x) repeatedly: f(2β
3)=2f(3)=2(1)=2, so f(6)=2. Next, f(2β
6)=2f(6)=2(2)=4, so f(12)=4. Continuing this pattern, f(2β
12)=2f(12)=8, f(48)=16, ... | 192 |
Let the roots of the quadratic equation x 2 +kx+5=0 be r 1 β and r 2 β . From Vieta's formulas, we know that r 1 β +r 2 β =βk and r 1 β r 2 β =5. The problem states that the roots differ by 61 β , which means β£r 1 β βr 2 β β£= 61 β . Squaring both sides gives (r 1 β βr 2 β ) 2 =61. We can express (r 1 β βr ... | 9 |
To solve the equation 6x+5 β 3x+5 β β = 3 5 β β , we first square both sides to eliminate the square roots: ( 6x+5 β 3x+5 β β ) 2 =( 3 5 β β ) 2 . This simplifies to 6x+5 3x+5 β = 9 5 β . Next, we cross-multiply to solve for x: 9(3x+5)=5(6x+5). Expanding both sides gives 27x+45=30x+25. Rearranging the terms... | 20/3 |
To find the area of a square, we first need to determine the length of its side, s. Since the points (β1,4) and (2,β3) are adjacent vertices, the distance between them is equal to the side length s. Using the distance formula, d= (x 2 β βx 1 β ) 2 +(y 2 β βy 1 β ) 2 β , we substitute the coordinates: s= (2β(β1)) 2... | 58 |
To find the greatest integer n such that n 2 β11n+24β€0, we first find the roots of the quadratic equation n 2 β11n+24=0 by factoring. We look for two numbers that multiply to 24 and add to -11, which are -3 and -8. Thus, (nβ3)(nβ8)β€0. This inequality holds when n is between the two roots, inclusive: 3β€nβ€8. The integers... | 8 |
The equation β£x+5β£=20 means that the expression x+5 can be either 20 or β20. We solve these two cases separately: 1) x+5=20βΉx=15. 2) x+5=β20βΉx=β25. To find the positive difference between these two solutions, we subtract the smaller from the larger: 15β(β25)=15+25=40. | 40 |
We are given three rules for the operation \#: (1) r \# 0 = r, (2) r \# s = s \# r, and (3) (r + 1) \# s = (r \# s) + s + 1. We want to find 11 \# 5. Let's explore the recurrence relation in (3). Starting from r \# s: 1 \# s = (0 \# s) + s + 1. By rule (2), 0 \# s = s \# 0, and by rule (1), s \# 0 = s. So, 1 \# s = s +... | 71 |
To find the sum of the possible values of x for the equation (x+2)(xβ3)=14, we first expand the left side: x 2 β3x+2xβ6=14, which simplifies to x 2 βxβ6=14. Subtracting 14 from both sides to set the quadratic to zero, we get x 2 βxβ20=0. This quadratic equation is in the form ax 2 +bx+c=0, where a=1, b=β1, and c=β20. A... | 1 |
To rationalize the denominator of the expression 2 β β1 1 β , we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of 2 β β1 is 2 β +1. This gives us ( 2 β β1)( 2 β +1) 1( 2 β +1) β . Applying the difference of squares formula, (aβb)(a+b)=a 2 βb 2 , to the denomi... | \sqrt{2}+1 |
In an arithmetic sequence, the n-th term is given by the formula a n β =a 1 β +(nβ1)d, where a 1 β is the first term and d is the common difference. We are given a 1 β =5 and a 13 β =29. Using the formula for the thirteenth term: 29=5+(13β1)d, which simplifies to 24=12d. Solving for d, we find d=2. Now, to find th... | 103 |
To find the sum of all solutions for the equation 20 4x β = x 5 β , we first simplify the left side: 20 4x β = 5 x β . The equation becomes 5 x β = x 5 β . Next, we cross-multiply: xβ
x=5β
5, which gives x 2 =25. Taking the square root of both sides, we find the two solutions: x=5 and x=β5. The question asks for th... | 0 |
The expression 2x 2 β12x+3 is a quadratic in the form ax 2 +bx+c, where a=2, b=β12, and c=3. Since a>0, the parabola opens upward, meaning the vertex represents the minimum value of the function. The x-coordinate of the vertex can be found using the formula x=βb/(2a). Substituting the given values, we get x=β(β12)/(2β2... | 3 |
Let n=βxβ be the greatest integer less than or equal to x. Then x can be written as n+f, where 0β€f<1. The equation becomes (n+f)n=70, or n 2 +nf=70. Since 0β€f<1, it follows that 0β€nf<n. Therefore, n 2 β€70<n 2 +n. We need to find an integer n such that n 2 is less than or equal to 70 and n(n+1) is greater than 70. Testi... | 8.75 |
Let the width of the rectangular garden be w and the length be l. We are given that the length is twice the width, so l=2w. The perimeter P of a rectangle is calculated using the formula P=2l+2w. Substituting the given perimeter and the relationship between length and width, we get 60=2(2w)+2w, which simplifies to 60=4... | 200 |
The function is given as f(x)=x 2 βx. To find the value of f(4), we substitute x=4 into the expression. This gives f(4)=4 2 β4. Calculating the square, we get 16β4. Subtracting 4 from 16 results in 12. | 12 |
For a triangle with side lengths a, b, and c to exist, the triangle inequality theorem states that the sum of any two sides must be greater than the third side: a+b>c, a+c>b, and b+c>a. Given the side lengths are 7, 10, and x 2 , we set up the following inequalities: 1) 7+10>x 2 βΉ17>x 2 , 2) 7+x 2 >10βΉx 2 >3, 3) 10+x 2... | 2, 3, 4 |
Let the integer be n. According to the problem, n 2 =n+182. This can be rewritten as a quadratic equation: n 2 βnβ182=0. For a quadratic equation in the form an 2 +bn+c=0, the sum of the roots is given by the formula βb/a. In this case, a=1 and b=β1. Therefore, the sum of all integers n that satisfy the equation is β(β... | 1 |
The height of the ball is given by the quadratic function h(t)=β16t 2 +64t+31. Since the coefficient of the t 2 term is negative (a=β16), the parabola opens downward, and the maximum height occurs at the vertex. The time t at which the vertex occurs is given by the formula t=βb/(2a). Here, a=β16 and b=64. Substituting ... | 95 |
First, calculate the total duration of the trip. From 9:40 a.m. to 12:00 p.m. is 2 hours and 20 minutes. From 12:00 p.m. to 1:20 p.m. is 1 hour and 20 minutes. Adding these together gives 3 hours and 40 minutes. Convert this time into hours: 40 minutes is 40/60=2/3 of an hour, so the total time is 3+2/3=11/3 hours. Ave... | 45 |
For the expression x 2 +bx+8 9x 3 +4x 2 +11x+7 β to have a domain of all real numbers, the denominator x 2 +bx+8 must never be equal to zero for any real x. This means the quadratic x 2 +bx+8 must have no real roots. For a quadratic ax 2 +bx+c to have no real roots, its discriminant D=b 2 β4ac must be strictly less th... | 5 |
The expression is 1. 3 0. 6 β . First, convert the repeating decimals to common fractions. Let x=0. 6 . Then 10x=6. 6 , and 10xβx=6. 6 β0. 6 , which gives 9x=6, so x= 9 6 β = 3 2 β . Let y=1. 3 . Then y=1+0. 3 . Since 0. 3 = 3 1 β , we have y=1+ 3 1 β = 3 4 β . The expression becomes y x β = 4/3 2/3 β = 3 2 β ... | 1/2 |
To find the midpoint between two points (x 1 β ,y 1 β ) and (x 2 β ,y 2 β ), I use the midpoint formula: M=( 2 x 1 β +x 2 β β , 2 y 1 β +y 2 β β ). Given the points (3,7) and (5,1), the x-coordinate is 2 3+5 β = 2 8 β =4 and the y-coordinate is 2 7+1 β = 2 8 β =4. Therefore, the coordinates are (4,4). | (4,4) |
The line is given in slope-intercept form y=mx+b. We are given the slope m=β7. To find b, we substitute the coordinates of the given point (3,0) into the equation: 0=β7(3)+b. This simplifies to 0=β21+b, which means b=21. The problem asks for the value of m+b. Substituting the values we found: m+b=β7+21=14. | 14 |
The equation of the circle is x 2 β6yβ3=βy 2 β4x. Rearranging terms to standard form, we get x 2 +4x+y 2 β6y=3. Completing the square for x: (x 2 +4x+4)β4. Completing the square for y: (y 2 β6y+9)β9. Substituting these back: (x+2) 2 β4+(yβ3) 2 β9=3. Simplifying gives (x+2) 2 +(yβ3) 2 =3+4+9=16. The standard form is (xβ... | 5 |
First, evaluate f(x) at x=4: f(4)=4 2 β2(4)+m=16β8+m=8+m. Next, evaluate g(x) at x=4: g(4)=4 2 β2(4)+4m=16β8+4m=8+4m. The problem states that 2f(4)=g(4), so we set up the equation 2(8+m)=8+4m. Expanding the left side gives 16+2m=8+4m. Subtracting 2m from both sides gives 16=8+2m. Subtracting 8 from both sides gives 8=2... | 4 |
The quadratic x 2 +txβ10 is factored as (x+a)(x+b), where a and b are integers. Expanding (x+a)(x+b) gives x 2 +(a+b)x+ab. By comparing coefficients, we have ab=β10 and a+b=t. Since a and b are integers, we find all possible pairs (a,b) such that ab=β10: (1,β10),(β1,10),(2,β5),(β2,5). The corresponding values for t=a+b... | 729 |
To factor the expression 58x 5 β203x 11 , I need to find the greatest common factor (GCF) of the two terms. First, look at the coefficients 58 and 203. 58=2Γ29. Testing 29 on 203: 203Γ·29=7. So, the GCF of the constants is 29. Next, look at the variable parts x 5 and x 11 . The GCF is the lowest power, which is x 5 . Th... | 29x^5(2-7x^6) |
We are asked to evaluate the expression (a 2 +b) 2 β(a 2 βb) 2 for a=4 and b=1. First, substitute the values: a 2 =4 2 =16. The expression becomes (16+1) 2 β(16β1) 2 , which is 17 2 β15 2 =289β225=64. Alternatively, notice this is a difference of squares: X 2 βY 2 =(XβY)(X+Y). Let X=a 2 +b and Y=a 2 βb. Then XβY=(a 2 +... | 64 |
The graph represents a function y=u(x). By observing the graph or the asy code provided (f1(x)=βx+3sin(xΟ/3)), we can check for symmetry. Let's test if the function is odd: u(βx)=β(βx)+3sin((βx)Ο/3)=xβ3sin(xΟ/3)=β(βx+3sin(xΟ/3))=βu(x). Since u(βx)=βu(x), the function u(x) is an odd function. Odd functions have the prop... | 0 |
Let the two numbers be x and y, where xβ₯y. We are given two equations: x+y=45 and xβy=3. To find the lesser number y, we can subtract the second equation from the first: (x+y)β(xβy)=45β3. This simplifies to 2y=42. Dividing by 2, we find y=21. To check, if y=21, then x=21+3=24, and 24+21=45, which is correct. | 21 |
We are given the equation m+ m 1 β =8. To find the value of m 2 + m 2 1 β +4, we first square the given equation: (m+ m 1 β ) 2 =8 2 . Expanding the left side gives m 2 +2(m)( m 1 β )+ m 2 1 β =64. Since mβ
m 1 β =1, this simplifies to m 2 +2+ m 2 1 β =64. Subtracting 2 from both sides, we find m 2 + m 2 1 β =6... | 66 |
To solve the quadratic equation 11x 2 β44xβ99=0 by completing the square, we first divide the entire equation by the leading coefficient, 11: x 2 β4xβ9=0. Next, we move the constant term to the other side: x 2 β4x=9. To complete the square for the expression x 2 β4x, we take half of the coefficient of x, which is β4/2=... | 11 |
The expression to evaluate is log 3 β 3 β 1 β . First, I rewrite the argument of the logarithm as a power of 3. The square root of 3 can be written as 3 1/2 . Therefore, 3 β 1 β = 3 1/2 1 β =3 β1/2 . Substituting this into the logarithm gives log 3 β (3 β1/2 ). Using the logarithm property log b β (b x )=x, we ... | -1/2 |
To calculate 31 2 from 30 2 , Emily uses the algebraic identity (n+1) 2 =n 2 +2n+1. For n=30, this is 31 2 =30 2 +2(30)+1=30 2 +61. To calculate 29 2 from 30 2 , we use the identity (nβ1) 2 =n 2 β2n+1, which can be written as (nβ1) 2 =n 2 β(2nβ1). For n=30, this is 29 2 =30 2 β(2(30)β1)=30 2 β(60β1)=30 2 β59. Therefore... | 59 |
To solve β£xβ7β£=β£x+1β£, we consider the possible cases for the signs of the expressions inside the absolute values. Case 1: xβ7=x+1, which simplifies to β7=1 (no solution). Case 2: xβ7=β(x+1), which simplifies to xβ7=βxβ1, then 2x=6, so x=3. Geometrically, this equation represents the set of points x that are equidistant... | 1 |
To solve β£yβ6β£+2y=9, we consider two cases for the absolute value. Case 1: yβ6β₯0, which means yβ₯6. In this case, β£yβ6β£=yβ6. The equation becomes (yβ6)+2y=9, or 3yβ6=9. Adding 6 gives 3y=15, so y=5. However, the condition for this case is yβ₯6, so y=5 is not a solution. Case 2: yβ6<0, which means y<6. In this case, β£yβ6β£... | 3 |
Let a and b be the roots of x 2 βmx+2=0. According to Vieta's formulas, we have a+b=m and ab=2. The second equation x 2 βpx+q=0 has roots r 1 β =a+ b 1 β and r 2 β =b+ a 1 β . Again, by Vieta's formulas, the product of the roots q is r 1 β r 2 β =(a+ b 1 β )(b+ a 1 β ). Expanding this product, we get q=ab+a( a ... | rac{9}{2} |
The function f(x)=3xβ2 is a linear function with a positive slope, meaning it is strictly increasing. Therefore, the composition g(x)=f(f(f(f(x)))) is also strictly increasing. To find the range of g on the domain [0,2], we simply need to evaluate g(0) and g(2). First, calculate the iterations for x=0: f(0)=3(0)β2=β2, ... | [-80, 82] |
To find the center of the circle, we rewrite the equation x 2 +y 2 =β2xβ10yβ16 in standard form (xβh) 2 +(yβk) 2 =r 2 . First, move all terms except the constant to the left: x 2 +2x+y 2 +10y=β16. Next, complete the square for x: x 2 +2x+1=(x+1) 2 . Complete the square for y: y 2 +10y+25=(y+5) 2 . Adding 1 and 25 to bo... | -6 |
Wanda needs to calculate the sum of the distances from P(4,2) to the vertices A(0,0), B(8,β1), and C(5,4). Using the distance formula d= (x 2 β βx 1 β ) 2 +(y 2 β βy 1 β ) 2 β : 1. PA= (4β0) 2 +(2β0) 2 β = 16+4 β = 20 β =2 5 β . 2. PB= (4β8) 2 +(2β(β1)) 2 β = (β4) 2 +3 2 β = 16+9 β = 25 β =5. 3. PC= (4β5) ... | 8 |
To simplify (3β2i) 2 , we expand the expression using the square of a binomial formula (aβb) 2 =a 2 β2ab+b 2 . Here, a=3 and b=2i. So, (3β2i) 2 =3 2 β2(3)(2i)+(2i) 2 =9β12i+4i 2 . Since i 2 =β1, we substitute it into the expression: 9β12i+4(β1)=9β12iβ4=5β12i. | 5-12i |
The expression given is the sum of two polynomials: (12d+13+14d 2 )+(2d+1). To simplify this, I need to combine like terms. First, identify the d 2 terms: there is only 14d 2 , so c=14. Next, identify the d terms: 12d+2d=14d, so a=14. Finally, identify the constant terms: 13+1=14, so b=14. The simplified expression is ... | 42 |
Let the sequence be a 1 β ,a 2 β ,a 3 β ,β¦. We are given that for n>1, each term is inversely proportional to the preceding term: a n β =k/a nβ1 β , where k is the constant of proportionality. We know a 1 β =2 and a 2 β =5. Using the formula for a 2 β , we have 5=k/2, so k=10. Now we can find the next few terms... | 5 |
The sum we want to maximize is S=fg+gh+hj+fj. We can factor this expression by grouping terms: S=g(f+h)+j(h+f)=(g+j)(f+h). We are given the set of values {5,6,7,8} to assign to the variables f,g,h,j. Let X=g+j and Y=f+h. We know that X+Y=5+6+7+8=26. To maximize the product XY where the sum X+Y is constant, we want X an... | 169 |
First, I will expand the left side of the equation (xβ5)(2x+9)=2x 2 +9xβ10xβ45=2x 2 βxβ45. The equation is then 2x 2 βxβ45=x 2 β13x+40. Moving all terms to the left side gives x 2 +12xβ85=0. Let p and q be the roots of this quadratic. By Vieta's formulas, p+q=β12 and pq=β85. We want to find (p+3)(q+3). Expanding this g... | -112 |
To solve the inequality n β β€ 4nβ6 β < 2n+5 β , we first ensure all expressions under the square roots are non-negative: nβ₯0, 4nβ6β₯0βΉnβ₯1.5, and 2n+5β₯0βΉnβ₯β2.5. Thus, n must be at least 2. Now, we square all parts of the inequality: nβ€4nβ6<2n+5. This gives two separate inequalities: 1) nβ€4nβ6βΉ6β€3nβΉnβ₯2. 2) 4nβ6<2n+5βΉ2n... | 4 |
To find the value of the expression (2x+5) 2 when x=3, we substitute 3 for x in the expression. First, calculate the value inside the parentheses: 2(3)+5=6+5=11. Then, square this result: 11 2 =121. | 121 |
The equation is 16 16 +16 16 +16 16 +16 16 =2 x . There are four identical terms on the left side, so we can write it as 4Γ16 16 =2 x . Expressing 4 and 16 as powers of 2, we get 4=2 2 and 16=2 4 . Substituting these into the equation gives 2 2 Γ(2 4 ) 16 =2 x . Using the power of a power rule (a m ) n =a mn , we have ... | 66 |
Let the two consecutive page numbers be n and n+1. The problem states that n(n+1)=18360, which simplifies to n 2 +nβ18360=0. To find n, we can approximate 18360 β . Since 130 2 =16900 and 140 2 =19600, the value of n must be between 130 and 140. Testing 135Γ136: 135Γ136=18360. Thus, n=135 and n+1=136. The sum of these... | 271 |
The equation is a quadratic of the form ax 2 +bx+c=0 with a=1,b=β5,c=5. The solutions are given by the quadratic formula x= 2a βbΒ± b 2 β4ac β β . Substituting the values, we get x= 2(1) 5Β± (β5) 2 β4(1)(5) β β = 2 5Β± 25β20 β β = 2 5Β± 5 β β . The two solutions are x 1 β = 2 5+ 5 β β and x 2 β = 2 5β 5 β β .... | \sqrt{5} |
To solve the equation 2 1 β + x 1 β = 6 5 β , we first isolate the term with x by subtracting 2 1 β from both sides: x 1 β = 6 5 β β 2 1 β . To subtract the fractions, we find a common denominator, which is 6: 2 1 β = 6 3 β . So, x 1 β = 6 5 β β 6 3 β = 6 2 β . Simplifying the fraction 6 2 β gives 3 1 β ... | 3 |
The graph shows a parabola f(x) with roots at x=1 and x=3. Since g(x)=βf(x), the graph of g(x) is the reflection of f(x) across the x-axis. Intersections occur where f(x)=g(x), which means f(x)=βf(x), or 2f(x)=0. This happens at the roots of f(x). Since there are 2 roots (x=1 and x=3), we have a=2. For h(x)=f(βx), this... | 21 |
The equation is x 2 β5x+5=9. To solve for x, we first set the equation to zero by subtracting 9 from both sides, resulting in x 2 β5xβ4=0. This is a quadratic equation of the form ax 2 +bx+c=0 where a=1,b=β5,c=β4. According to Vieta's formulas, for a quadratic equation ax 2 +bx+c=0, the sum of the roots is given by βb/... | 5 |
The expression is 1+ 3 β +1 1 β 1 β . First, simplify the denominator: 1+ 3 β +1 1 β = 3 β +1 3 β +1+1 β = 3 β +1 3 β +2 β . Now, the expression becomes 3 β +1 3 β +2 β 1 β = 3 β +2 3 β +1 β . To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, w... | \sqrt{3}-1 |
The sequence is arithmetic with the first term a 1 β =1. The common difference d is the difference between consecutive terms: d=10β1=9 (and 19β10=9). The formula for the n-th term of an arithmetic sequence is a n β =a 1 β +(nβ1)d. To find the 21st term (n=21), we substitute the values: a 21 β =1+(21β1)9=1+(20)9=1+1... | 181 |
The inequality (n+3)(nβ7)β€0 is satisfied when the product of two factors is less than or equal to zero. This happens when the factors have opposite signs or when one is zero. The roots of the quadratic expression (n+3)(nβ7) are n=β3 and n=7. Since the leading coefficient of n 2 β4nβ21 is positive, the parabola opens up... | 11 |
The problem asks for the value of aβb+c for the quadratic function y=ax 2 +bx+c. This value is equivalent to the function f(x) evaluated at x=β1, since f(β1)=a(β1) 2 +b(β1)+c=aβb+c. By looking at the provided graph, we can determine the coordinates of points on the parabola. The vertex appears to be at (β1,β2). To conf... | -2 |
The expression is x 4 +6 x 8 +12x 4 +36 β . Notice that the numerator is a perfect square trinomial. Let y=x 4 . Then the expression becomes y+6 y 2 +12y+36 β . The numerator y 2 +12y+36 factors into (y+6) 2 . Thus, the expression simplifies to y+6 (y+6) 2 β =y+6. Substituting y=x 4 back in, we get x 4 +6. Now, we e... | 631 |
We are given the equations xβy=15 and xy=4. To find the value of x 2 +y 2 , we can use the algebraic identity (xβy) 2 =x 2 β2xy+y 2 . Substituting the known values into this identity, we get (15) 2 =x 2 β2(4)+y 2 . This simplifies to 225=x 2 β8+y 2 . To isolate x 2 +y 2 , we add 8 to both sides of the equation: 225+8=x... | 233 |
First, we rewrite the equation of the circle x 2 +y 2 =β4x+6yβ12 in standard form (xβh) 2 +(yβk) 2 =r 2 to find its center (h,k). Moving the x and y terms to the left side gives x 2 +4x+y 2 β6y=β12. Completing the square for x: x 2 +4x+4=(x+2) 2 . Completing the square for y: y 2 β6y+9=(yβ3) 2 . Adding 4 and 9 to both ... | 5 |
A geometric sequence is defined by the formula a n β =a 1 β β
r nβ1 . We are given the first term a 1 β =3 and the fourth term a 4 β =192. Substituting these values into the formula for n=4, we get 192=3β
r 4β1 , which simplifies to 192=3β
r 3 . Dividing both sides by 3, we find r 3 =64. Taking the cube root of both s... | 48 |
A vertical asymptote of a rational function y= g(x) f(x) β occurs where the denominator g(x) is equal to zero, provided that the numerator f(x) is not zero at that same point (which would indicate a hole instead). For the function y= 5xβ7 x+2 β , the denominator is 5xβ7. We find the value of x that makes the denomina... | rac{7}{5} |
A geometric sequence follows the formula a n β =a 1 β β
r nβ1 . We are given the first term a 1 β =2 and the fifth term a 5 β =162. Substituting these values into the formula for n=5, we have 162=2β
r 5β1 , which simplifies to 162=2β
r 4 . Dividing both sides by 2, we get r 4 =81. Since the sequence consists of positi... | 486 |
Let the center of the circle be (h,0) since it lies on the x-axis. The radius r is the distance from the center to any point on the circle. Thus, the squared distance from (h,0) to (0,4) must equal the squared distance from (h,0) to (1,3). Using the distance formula: (0βh) 2 +(4β0) 2 =(1βh) 2 +(3β0) 2 . This simplifies... | 5 |
Let the x-coordinates of the vertices of the 33-gon P 1 β be x 1 β ,x 2 β ,β¦,x 33 β . We are given that β i=1 33 β x i β =99. The vertices of P 2 β are the midpoints of the sides of P 1 β . Let the x-coordinates of the vertices of P 2 β be y 1 β ,y 2 β ,β¦,y 33 β . These are given by y i β = 2 x i β +x i+1... | 99 |
Let u= x+2 1 β . If x>β2, then x+2>0, so u>0. The range of u for xβ(β2,β) is (0,β). In this case, f(x)=βuβ. For u>0, the ceiling function βuβ takes on all positive integer values {1,2,3,β¦}. If x<β2, then x+2<0, so u<0. The range of u for xβ(ββ,β2) is (ββ,0). In this case, f(x)=βuβ. For u<0, the floor function βuβ take... | 0 |
The goal is to find the value of f β1 (f β1 (f β1 (1))) using the given table for f(x). The table defines f(x) for x=1,2,3,4,5. To find f β1 (y), we look for the value x such that f(x)=y. First, find f β1 (1): from the table, f(2)=1, so f β1 (1)=2. Next, find f β1 (f β1 (1))=f β1 (2): from the table, f(5)=2, so f β1 (2... | 3 |
The expression is 21 β 21 β . To rationalize the denominator, we multiply both the numerator and the denominator by 21 β . This gives 21 β β
21 β 21β
21 β β . The denominator 21 β β
21 β simplifies to 21. So the expression becomes 21 21 21 β β . The factor of 21 in the numerator and denominator cancels out, ... | \sqrt{21} |
The expression is y=β£x+7β£ββ£xβ2β£. We can analyze this function by examining the critical points where the expressions inside the absolute values change sign, which are x=β7 and x=2. This divides the number line into three intervals: 1) x<β7: Both x+7 and xβ2 are negative. So y=β(x+7)β(β(xβ2))=βxβ7+xβ2=β9. 2) β7β€xβ€2: x+7... | [-9, 9] |
The problem states that a 2 varies inversely with b 3 . This relationship can be expressed by the equation a 2 = b 3 k β , where k is a constant of proportionality. We are given that a=7 when b=3. Substituting these values into the equation, we get 7 2 = 3 3 k β , which simplifies to 49= 27 k β . Solving for k, we f... | rac{49}{8} |
To solve the equation 5 x+4 =125 x , we first express both sides with the same base. Since 125=5 3 , we can rewrite the right side as (5 3 ) x =5 3x . The equation then becomes 5 x+4 =5 3x . Since the bases are equal and positive (and not equal to 1), we can equate the exponents: x+4=3x. Solving for x, we subtract x fr... | 2 |
To solve the equation (xβ4) 3 =( 8 1 β ) β1 , first simplify the right side. The expression ( 8 1 β ) β1 is equal to the reciprocal of 8 1 β , which is 8. The equation now reads (xβ4) 3 =8. To solve for x, take the cube root of both sides. Since 8=2 3 , the cube root of 8 is 2. This gives the linear equation xβ4=2. ... | 6 |
In an arithmetic sequence, the difference between consecutive terms is constant. Given the terms 1 2 , x 2 , and 3 2 , let the common difference be d. We have a 1 β =1, a 2 β =x 2 , and a 3 β =9. The condition for an arithmetic sequence is a 2 β βa 1 β =a 3 β βa 2 β . Substituting the values, we get x 2 β1=9βx 2... | \sqrt{5} |
The temperature is given by the quadratic function f(t)=βt 2 +12t+50. We need to find the largest value of t for which the temperature is exactly 77 degrees. Setting the expression equal to 77, we have βt 2 +12t+50=77. Rearranging the equation into standard quadratic form at 2 +bt+c=0, we subtract 77 from both sides to... | 9 |
We start with the expression 2x+8x 2 +9β(4β2xβ8x 2 ). To simplify it, we first distribute the negative sign into the parentheses: 2x+8x 2 +9β4+2x+8x 2 . Next, we combine the like terms: the x 2 terms are 8x 2 +8x 2 =16x 2 , the x terms are 2x+2x=4x, and the constant terms are 9β4=5. Summing these up, the simplified exp... | 16x^2+4x+5 |
We are given a system of two linear equations with variables x, y, and a constant a: (1) 3x+y=a and (2) 2x+5y=2a. We are told that x=2 is part of the solution. First, substitute x=2 into both equations. Equation (1) becomes 3(2)+y=a, which simplifies to 6+y=a. Equation (2) becomes 2(2)+5y=2a, which simplifies to 4+5y=2... | rac{26}{3} |
We are given a system of four linear equations: (1) aβb+c=5, (2) bβc+d=6, (3) cβd+a=3, and (4) dβa+b=2. We want to find the value of a+b+c+d. Let's try adding all four equations together. The left side is (aβb+c)+(bβc+d)+(cβd+a)+(dβa+b). Combining like terms: the a terms are a+aβa=a; the b terms are βb+b+b=b; the c ter... | 16 |
To solve the inequality x 2 +5x<6, we first rewrite it in standard quadratic form by subtracting 6 from both sides, yielding x 2 +5xβ6<0. Next, we factor the quadratic expression by finding two numbers that multiply to β6 and add to 5; these are 6 and β1. Thus, the inequality becomes (x+6)(xβ1)<0. We then identify the ... | (-6, 1) |
To solve for A and B, we first multiply the entire equation by (xβ5) to clear the denominator, assuming x ξ =5. This gives: A+B(x+1)(xβ5)=β3x 2 +12x+22. Expanding the left side: A+B(x 2 β4xβ5)=β3x 2 +12x+22, which simplifies to Bx 2 β4Bx+(Aβ5B)=β3x 2 +12x+22. By comparing the coefficients of the terms on both sides: 1)... | 4 |
The expression t 2 β49 is a difference of two squares because both t 2 and 49 are perfect squares (t 2 =(t) 2 and 49=7 2 ). The formula for factoring a difference of squares is a 2 βb 2 =(aβb)(a+b). Substituting a=t and b=7 into the formula gives (tβ7)(t+7). | (t-7)(t+7) |
We are given two equations: (1) x 2 +y 2 =90 and (2) xy=27. We want to find the greatest possible value of x+y. We know the identity (x+y) 2 =x 2 +2xy+y 2 . Substituting the given values into this identity, we get (x+y) 2 =(x 2 +y 2 )+2(xy)=90+2(27)=90+54=144. Taking the square root of both sides, we find x+y=Β± 144 β ... | 12 |
We are given a quadratic equation x 2 +bxβ28=0 and told that x=β7 is a solution. By the definition of a solution, substituting x=β7 into the equation must satisfy it. Plugging in the value, we get (β7) 2 +b(β7)β28=0. Simplifying the terms, we have 49β7bβ28=0. Combining the constant terms 49β28 gives 21, so the equation... | 3 |
The expression to evaluate is ββ 4 49 β β β. First, simplify the square root. Since 4 49 β =( 2 7 β ) 2 , the square root 4 49 β β equals 2 7 β , which is 3.5. The expression then becomes ββ3.5β. The ceiling function βxβ is defined as the smallest integer greater than or equal to x. Looking at a number line , we... | -3 |
First, I need to calculate the total amount of flour Jessica uses each day. She makes 30 pie crusts, and each uses 1/6 cup of flour. The total flour is 30Γ(1/6)=30/6=5 cups. Next, Jessica wants to make 20 larger pie crusts using this same total of 5 cups of flour. To find the amount of flour per new crust, I divide the... | 1/4 |
To find the values of a and b for the line y=ax+b, we substitute the given points into the equation. For (4,5), we get 5=4a+b. For (8,17), we get 17=8a+b. This is a system of two linear equations. Subtracting the first equation from the second: (17β5)=(8aβ4a)+(bβb), which simplifies to 12=4a, so a=3. Substituting a=3 b... | 10 |
We start with the second equation and expand it: (a m xβa n )(a p yβa 2 )=a m x(a p y)βa m x(a 2 )βa n (a p y)+a n (a 2 )=a m+p xyβa m+2 xβa n+p y+a n+2 . We are given that this is equivalent to a 7 xyβa 6 yβa 5 x=a 4 b 4 βa 4 , which can be rearranged as a 7 xyβa 5 xβa 6 y+a 4 =a 4 b 4 . Comparing the two expanded for... | 24 |
The expression is (2+1)(2 2 +1)(2 4 +1). This is a product of terms in the form (2 2 n +1). To solve this efficiently, we can multiply the expression by (2β1), which equals 1 and doesn't change the value. Using the difference of squares identity (aβb)(a+b)=a 2 βb 2 : First, (2β1)(2+1)=2 2 β1. Next, (2 2 β1)(2 2 +1)=2 4... | 255 |
We are given three equations: (1) x+ y 1 β =4, (2) y+ z 1 β =1, and (3) z+ x 1 β = 3 7 β . To find xyz, we multiply the three equations together: (x+ y 1 β )(y+ z 1 β )(z+ x 1 β )=4Γ1Γ 3 7 β = 3 28 β . Expanding the left side: (xy+ z x β +1+ yz 1 β )(z+ x 1 β )=xyz+y+x+ z 1 β +z+ x 1 β + y 1 β + xyz 1 β... | 1 |
To simplify the expression 3x+7x 2 +5β(2β3xβ7x 2 ), I need to distribute the negative sign into the parentheses. This changes the signs of all terms inside: 3x+7x 2 +5β2+3x+7x 2 . Next, I group the like terms together. The x 2 terms are 7x 2 +7x 2 , which equals 14x 2 . The x terms are 3x+3x, which equals 6x. The const... | 14x^2+6x+3 |
To evaluate the expression β 140 β β, I first need to estimate the value of 140 β . I can do this by finding the two perfect squares that 140 falls between. I know that 11 2 =121 and 12 2 =144. Since 121<140<144, it follows that 121 β < 140 β < 144 β , which simplifies to 11< 140 β <12. The ceiling function βxβ i... | 12 |
End of preview. Expand in Data Studio
README.md exists but content is empty.
- Downloads last month
- 7