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Algebra | input 5 | For the piecewise function to be continuous, the cases must "meet" at $2$ and $-2$. For example, $ax+3$ and $x-5$ must be equal when $x=2$. This implies $a(2)+3=2-5$, which we solve to get $2a=-6 \Rightarrow a=-3$. Similarly, $x-5$ and $2x-b$ must be equal when $x=-2$. Substituting, we get $-2-5=2(-2)-b$, which implies $b=3$. So $a+b=-3+3=\boxed{0}$. | Let \[f(x) = \left\{
\begin{array}{cl} ax+3, &\text{ if }x>2, \\
x-5 &\text{ if } -2 \le x \le 2, \\
2x-b &\text{ if } x <-2.
\end{array}
\right.\]Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper). |
Algebra | input 3 | Adding real parts and imaginary parts separately, we have $(2-(-4)+0+2)+(1+0-1+4)i=\boxed{8+4i}$. | If $A=2+i$, $O=-4$, $P=-i$, and $S=2+4i$, find $A-O+P+S$. |
Algebra | input 3 | Let one pair of parallel sides have length $x$ and the other pair of parallel sides have length $12-x$. This means that the perimeter of the rectangle is $x+x+12-x+12-x=24$ as the instruction states. The area of this rectangle is $12x-x^2$. Completing the square results in $-(x-6)^2+36\le 36$ since $(x-6)^2\ge 0$, so the maximum area of $\boxed{36}$ is obtained when the rectangle is a square of side length 6 inches. | The perimeter of a rectangle is 24 inches. What is the number of square inches in the maximum possible area for this rectangle? |
Algebra | input 3 | The given expression is undefined when the denominator is zero. Thus, we want to find the sum of the zeros $y$ to the quadratic $y^2-5y+4$. Since for a quadratic with the equation $ax^2+bx+c=0$, the sum of the outputs is $-b/a$, the sum of the zeros of the quadratic $y^2-5y+4$ is $5/1=\boxed{5}$. | What is the sum of all values of $y$ for which the expression $\frac{y+6}{y^2-5y+4}$ is undefined? |
Algebra | input 3 | Because the question only asks for the value of $q$, we can begin by eliminating $p$. In order to do this, we multiply the first equation by 4 and the second equation by 3, giving us a system of two equations that both have 12 as the coefficient of $p$ \begin{align*} 12p+16q&=32
\\ 12p+9q&=39
\end{align*}From here, we can just subtract the second equation from the first. This gives us $(12p+16q)-(12p+9q)=32-(39)$, which simplifies to $7q=-7$ or $q=\boxed{-1}$. | If $3p+4q=8$ and $4p+3q=13$, what is $q$ equal to? |
Algebra | input 2 | We use the distance formula to find that the distance is \[\sqrt{(-5 -7)^2 + (-2-3)^2} = \!\sqrt{144 + 25} = \boxed{13}.\] | Find the distance between the points $(-5,-2)$ and $(7,3)$. |
Algebra | input 5 | We see that $-2$ is not in the range of $f(x) = x^2 + bx + 2$ if and only if the equation $x^2 + bx + 2 = -2$ has no real roots. We can re-write this equation as $x^2 + bx + 4 = 0$. The discriminant of this quadratic is $b^2 - 4 \cdot 4 = b^2 - 16$. The quadratic has no real roots if and only if the discriminant is negative, so $b^2 - 16 < 0$, or $b^2 < 16$. The set of values of $b$ that satisfy this inequality is $b \in \boxed{(-4,4)}$. | For what values of $b$ is $-2$ not in the range of the function $f(x)=x^2+bx+2$? Express your answer in interval notation. |
Algebra | input 1 | Let $d$ equal the cost of a 70 mile taxi ride. Since we know that Ann was charged $120 dollars a 50 mile taxi ride, we can set up the proportion $\frac{120}{50}=\frac{d}{70}$. If we solve for $d$ by multiplying both sides by 70, we find that $d=\left(\frac{120}{50}\right)(70)=\boxed{168}$ dollars. | After traveling 50 miles by taxi, Ann is charged a fare of $\$120$. Assuming the taxi fare is directly proportional to distance traveled, how much would Ann be charged (in dollars) if she had traveled 70 miles? |
Algebra | input 1 | If the number is $x$, we can set up the equation $\frac{16}{x}=\frac{64}{100}$. We divide both sides by $4$ to get $\frac{1}{x}=\frac{4}{100}=\frac{1}{25}$, so $x=\boxed{25}$. | Sixteen is 64$\%$ of what number? |
Algebra | input 2 | Because $3\ge -2,$ we use the second case to determine that $f(3) = 5-2(3) = \boxed{-1}.$ | Let \[f(x) =
\begin{cases}
2x + 9 &\text{if }x<-2, \\
5-2x&\text{if }x\ge -2.
\end{cases}
\]Find $f(3).$ |