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Using Formulas 3.4 p. 143. Using Formulas to Solve Problems A formula is an equation that shows a relationship between values that are represented by.

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Presentation on theme: "Using Formulas 3.4 p. 143. Using Formulas to Solve Problems A formula is an equation that shows a relationship between values that are represented by."— Presentation transcript:

1 Using Formulas 3.4 p. 143

2 Using Formulas to Solve Problems A formula is an equation that shows a relationship between values that are represented by variables. The formula, d = rt, shows a relation between the distance traveled, the rate of speed, and the time traveled at this rate of speed. Our objective is to apply formulas to solve problems! A formula is an equation that shows a relationship between values that are represented by variables. The formula, d = rt, shows a relation between the distance traveled, the rate of speed, and the time traveled at this rate of speed. Our objective is to apply formulas to solve problems!

3 Example Problem p. 143 Suppose you travel 162 miles in 3 hours. We have to go back to the familiar process of WRITE, SUBSTITUTE, and SOLVE. Write the formula that applies to the problem d = rt Substitute the info we have. 162 = r (3) Solve, using equality properties 162 = 3r 3 54 = r The rate of speed was 54 mph.

4 When you see this calculator, it means you are allowed to use a calculator on the problem. You must follow the rules for calculator lessons. These are: WRITE each step BEFORE solving Show all steps Label all work

5 The Distance Formula If you drove for 9 ¾ hours and traveled a total of 273 miles, how fast were you going? distance = rate of speed(time)d = rt Write Substitute what you have. 273 = (r)9.75 Solve 273 = (r) 9.75 9.75 28 mph=rate of speed

6 The Cricket Problem p. 143Someone with way too much time on their hands found they could estimate the temperature by counting the frequency of cricket chirps. F = + 37 If a cricket was chirping about 96 times per minute, you could find the temperature by substituting and solving this formula. If a cricket was chirping about 96 times per minute, I would step on it. F = + 37 96 4 F = 24 + 37 F = 61 o

7 Perimeter Formulas Perimeter is the distance around a shape. Rectangular perimeter can be found by these formulas: P = 2l + 2w or P = 2(l + w) PLEASE notice that the first one is just the second one after the 2 has been distributed!! P = 2(18.5) + 2(12.5) P = 37 + 25 P = 62 ft. P = 2(18.5 + 12.5) P = 2(31) P = 62 ft. 18.5 ft. 12.5 ft.

8 Perimeter Formulas Perimeter is the distance around a shape. Rectangular perimeter can be found by these formulas: P = 2l + 2w or P = 2(l + w) Choose which formula you want.. Follow the complete process P = 2(27.3) + 2(16.8) P = 54.6 + 33.6 P = 88.2 cm P = 2(27.3 + 16.8) P = 2(44.1) P = 88.2 cm P = 2(17.4) + 2(8.6) P = 34.8 + 17.2 P = 52 ft. P = 2(17.4 + 8.6) P = 2(26) P = 52 ft. Would you ever use this?

9 Formula Practice p. 145 (Even 2-16) 2) d = rt 2730 = r (9.75 )___________ 9.75 280 mph= r Notice how you can manipulate the formula to find each part. d = rt To find the time, We would divide both sides by the rate. d = rt r r To find the distance, we would just multiply the rate(time). d = rt

10 4) d = r t 10.2 ft =.5ft (t) _______________.5.5 20.4 hours = time 6) F = + 37 F = 80 + 37 4 F = 20 + 37 F = 57 degrees 8) F = + 37 F = 64 + 37 4 F = 16 + 37 F = 53 degrees

11 10) P = 2(l + w) P = 2( 7.3 + 6.2) P = 2( 13.5) P = 27 m F = 1.8C + 32 Now that the cricket is dead we will use a formula to convert C to F. 12) F = 1.8C + 32 F = 1.8(-4) + 32 F = -7.2 + 32 F = 24.8 14) F = 1.8C + 32 F = 1.8(56) + 32 F = 100.8 + 32 F = 132.8

12 Challenge There is a type of formula that is very similar to the distance formula. It’s called the work formula. Work completed = rate of work (time) We will work two examples of these problems. Keep an open mind……

13 Frank can cut a lawn in 2 hours. His brother Jeff can cut the same lawn in 3 hours. How long will it take them if they cut the lawn at the same time? We are asked to find the time, in hours, it will take both people to finish the lawn. Since we are looking for time in hours, Frank can cut ½ the yard in an hour and Jeff can cut 1/3 the yard in an hour.

14 The “pipe problems” are problems you will see quite often in algebra…… One pipe can fill a tank in 5 hours. Another pipe can fill the same tank in 3 hours. How long will it take to fill the tank with both pipes? Oh, no!!!! Only the pipe that fills the tank in 3 hours is working! On top of that, the drain is open. The drain can let all the water out in 9 hours. How long will it take to fill the tank now??????

15 What Did We Accomplish? Did you use formulas to solve problems? Hopefully, you remembered to follow the process of write, substitute, and solve. Hopefully, you remembered that you were using equality properties to solve these problems!


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