1 / 10

Warm Up 1. What is 35 increased by 8%? 2. What is the percent of decrease from 144 to 120?

2. 3. 16 %. Warm Up 1. What is 35 increased by 8%? 2. What is the percent of decrease from 144 to 120? 3. What is 1500 decreased by 75%? 4. What is the percent of increase from 0.32 to 0.64?. 37.8. 375. 100%. Derive formulas for P, r & t. I = P  r  t Memorize.

zelia
Download Presentation

Warm Up 1. What is 35 increased by 8%? 2. What is the percent of decrease from 144 to 120?

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 2 3 16 % Warm Up 1.What is 35 increased by 8%? 2. What is the percent of decrease from 144 to 120? 3. What is 1500 decreased by 75%? 4. What is the percent of increase from 0.32 to 0.64? 37.8 375 100%

  2. Derive formulas for P, r & t I = Pr tMemorize P = I/ (r t)Divide both sides by r*t r = I/(P t)Divide both sides by P*t t = I/(Pr)Divide both sides by P*r

  3. Solving for Interest Rate Mr. Johnson borrowed $8000 for 4 years to make home improvements. If he repaid a total of $10,320, at what interest rate did he borrow the money? P + I = AUse the formula. 8000 + I = 10,320 Substitute. I = 10,320 – 8000 = 2320 Subtract 8000 from both sides. He paid $2320 in interest. Use the amount of interest to find the interest rate.

  4. 1 4 2320 = rDivide both sides by 32,000. 32,000 Mr. Johnson borrowed the money at an annual rate of 7.25%, or 7 %. Example Continued I = Pr tUse the formula. 2320 = 8000 r4 Substitute. 2320 = 32,000 rSimplify. 0.0725 = r

  5. Example: Solving for Time Mr. Mogi borrowed $9000 at an interest rate of 12% to make home improvements. If he repaid a total of $20,000, how long did he borrow the money? P + I = AUse the formula. 9000 + I = 20,000 Substitute. I = 20,000 – 9000 = 11,000 Subtract 9000 from both sides. He paid $11,000 in interest. Use the amount of interest to find the time in years.

  6. 11,000 = tDivide both sides by 1080. 1080 Time Example Continued I = PrtUse the formula. 11,000 = 9000  .12 tSubstitute. 11,000 = 1080 tSimplify. 10.18 = t Mr. Mogi borrowed the money for over 10 years.

  7. Example 2: Solve for Time Nancy invested $6000 in a bond at a yearly rate of 3%. She earned $450 in interest. How long was the money invested? I = P r  t Use the formula. 450 = 6000  0.03 tSubstitute values into the equation. 450 = 180t 2.5 = tSolve for t. The money was invested for 2.5 years, or 2 years and 6 months. Guided Practice 2

  8. Example 3 TJ invested $4000 in a bond at a yearly rate of 2%. He earned $200 in interest. How long was the money invested? I = PrtUse the formula. 200 = 4000  0.02 t Substitute values into the equation. 200 = 80t 2.5 = tSolve for t. The money was invested for 2.5 years, or 2 years and 6 months. Guided Practice

  9. Homework Instructions Complete ‘I=PxRxT’ Work Sheet Back Side Chart #1-7

  10. Derive formulas for P, r & t I = Pr tMemorize P = I/ (r t)Divide both sides by r*t r = I/(P t)Divide both sides by P*t t = I/(Pr)Divide both sides by P*r

More Related