In one sentence, describe what the following article is about: Work through a return row, using either the Tunisian simple stitch or Tunisian double crochet. Wait until only two loops remain on your hook. Attach the new color to the tip of your hook, then draw this new color through both loops already on your hook. To attach the new yarn, tie it to the tip of your crochet hook with a standard slipknot. When you finish this step, you should have one loop of your new color on your hook and none of the old color on your hook. Work your next forward row of Tunisian simple stitches or Tunisian double crochet as you usually would, using the new yarn color instead of the old. Continue along with forward and reverse passes as usual until you have all you need for this color yarn. You can then change colors again, using the same technique, or finish off your work.
Summary: Reach the end of a return row. Draw the new color through both loops. Work your forward row as usual.

In one sentence, describe what the following article is about: For periodic compounding with multiple compounding per year, the future value is given by FV = PV (1 + r/n)^nT, where FV = future value, PV = present value, r = growth rate, T = time, and n = number of compounding periods per year. For continuous compounding, n approaches infinity. Using the definition of e = lim (1 + 1/n)^n as n approaches infinity, the expression becomes FV = PV e^(rT). If money has doubled, FV = 2*PV, so 2PV = PV e^(rT), or 2 = e^(rT), assuming the present value is not zero. Solve for T by taking natural logs on both sides, and rearranging, to get T = ln(2)/r = 69.3/R (where R = 100r to express the growth rate as a percentage). This is the rule of 69.3. For continuous compounding, 69.3 (or approximately 69) gives more accurate results, since ln(2) is approximately 69.3%, and R * T = ln(2), where R = growth (or decay) rate, T = the doubling (or halving) time, and ln(2) is the natural log of 2. 70 may also be used as an approximation for continuous or daily (which is close to continuous) compounding, for ease of calculation. These variations are known as rule of 69.3, rule of 69, or rule of 70. A similar accuracy adjustment for the rule of 69.3 is used for high rates with daily compounding: T = (69.3 + R/3) / R. The Eckart-McHale second order rule, or E-M rule, gives a multiplicative correction to the Rule of 69.3 or 70 (but not 72), for better accuracy for higher interest rate ranges. To compute the E-M approximation, multiply the Rule of 69.3 (or 70) result by 200/(200-R), i.e., T = (69.3/R) * (200/(200-R)). For example, if the interest rate is 18%, the Rule of 69.3 says t = 3.85 years. The E-M Rule multiplies this by 200/(200-18), giving a doubling time of 4.23 years, which better approximates the actual doubling time 4.19 years at this rate. The third-order Pade approximant gives even better approximation, using the correction factor (600 + 4R) / (600 + R), i.e., T = (69.3/R) * ((600 + 4R) / (600 + R)). If the interest rate is 18%, the third-order Pade approximant gives T = 4.19 years. To estimate doubling time for higher rates, adjust 72 by adding 1 for every 3 percentages greater than 8%. That is, T = [72 + (R - 8%)/3] / R. For example, if the interest rate is 32%, the time it takes to double a given amount of money is T = [72 + (32 - 8)/3] / 32 = 2.5 years. Note that 80 is used here instead of 72, which would have given 2.25 years for the doubling time. Here is a table giving the number of years it takes to double any given amount of money at various interest rates, and comparing the approximation with various rules:
Summary: Understand how the derivation works for continuous compounding.

In one sentence, describe what the following article is about: Start solving this just as you would normally, by comparing the divisor to the first digit of the dividend. Calculate the number of times the divisor goes into this digit, then write this number above that digit. Example: We're trying to fit 12 into 30. Compare 12 to the first digit of the divisor, 3. Since 12 is larger than 3, it goes into it 0 times. Write 0 above the 3, on the answer line. Write the product (the answer to the multiplication problem) down below the dividend. Write it directly below the first digit of the dividend, since this is the digit you just looked at. Example: Since 0 x 12 = 0, write 0 underneath the 3. Subtract the product you just found from the digit directly above it. Write the answer on a new line below. Example: 3 - 0 = 3, so write 3 directly below the 0. Bring the next digit of the dividend down next to the number you just wrote. Example: Our dividend is 30. We've already looked at the 3, so the next digit to bring down is 0. Bring this down next to your 3 to make 30. Now repeat the first step of this section to find the second digit of your answer. This time, compare the divisor to the number you just wrote down on the lowest line. Example:' How many times does 12 fit into 30? The closest we can get is 2, since 12 x 2 = 24. Write 2 in the second spot of the answer line. If you're not sure what the answer is, try some multiplication problems until you find the largest answer that fits. For example, if it seems like 3 is about write, multiply out 12 x 3 and you'll get 36. This is too big, since we're trying to fit within 30. Try the next one down, 12 x 2 = 24. This does fit, so 2 is the correct answer. This is the same long division process used above, and for any long division problem: Multiply the new digit on your answer line by the divisor: 2 x 12 = 24. Write the product on a new line below your dividend: Write 24 directly underneath 30. Subtract the lowest line from the one above it: 30 - 24 = 6, so write 6 on a new line underneath. If there's still another digit left in your dividend, bring it down and continue solving the problem the same way. If you've reached the end of the answer line, go to the next step. Example: We just wrote 2 at the end of the answer line. Go to the next step. If the numbers divided evenly, your last subtraction problem has "0" as the answer. That means you're done, and you have a whole number as the answer to your problem. But if you've reached the end of the answer line and there's still something left to divide, you'll need to extend the dividend by adding a decimal point followed by a 0. Remember, this does not change the value of the number. Example: We're at the end of the answer line but the answer to our last subtraction problem is "6." Extend the "30" under the long division symbol by adding ".0" to the end. Write a decimal point at the same spot on the answer line as well, but don't write anything after it yet. The only difference here is that you must bring the decimal point up to the same spot on the answer line. Once you've done that, finding the remaining digits of the answer is exactly the same. Example: Bring down the new 0 down to the last line to make "60." Since 12 goes into 60 exactly 5 times, write 5 as the last digit on our answer line. Don't forget that we put a decimal on our answer line, so 2.5 is the final answer to our problem.
Summary:
Find the first digit of the answer. Multiply that digit by the divisor. Subtract to find what's left over. Bring down the next digit. Try to fit the divisor into the new number. Repeat the steps above to find the next number. Continue until you reach the end of the answer line. Add a decimal to extend the dividend if necessary. Repeat the same steps to find the next digit.