(For help, go to Lessons 1-2, 1-4 and 1-6.)

Evaluate each formula for the values given.

1. distance: d = rt, when r = 60 mi/h and t = 3 h

2. perimeter of a rectangle: P = 2 + 2w, when = 11 cm, and w = 5 cm

3. area of a triangle: A = bh, when b = 8 m and h = 7 m12

FormulasALGEBRA 1 LESSON 2-6

2-6

Solutions

1. d = rt = 60(3) = 180 mi

2. P = 2 + 2w = 2(11) + 2(5) = 22 + 10 = 32 cm

3. A = bh = (8)(7) = 4(7) = 28 m212

12

FormulasALGEBRA 1 LESSON 2-6

2-6

Solve the formula for the volume of a rectangular prism V = wh for width w in terms of its volume V, length , and its height h.

V = wh

= wh Simplify.V

FormulasALGEBRA 1 LESSON 2-6

2-6

= Divide each side by , = 0.V wh /

= Divide each side by h, h = 0, to get w alone on one side of the equation.

V h

wh h /

= w Simplify. V h

Solve y = 4x – 3 for x.

y + 3 = 4x – 3 + 3 Add 3 to each side.

y + 3 = 4x Simplify.

FormulasALGEBRA 1 LESSON 2-6

2-6

= Divide each side by 4.y + 34

4x4

= x Simplify.y + 34

Solve z – br = p for b in terms of z, r, and p.

z – br – z = p – z Subtract z from each side.

–br = p – z Combine like terms.

FormulasALGEBRA 1 LESSON 2-6

2-6

= Divide each side by –r.–br–r

p – z–r

b = – Simplify.p – z–r

The formula K = C + 273.15 gives the Kelvin temperature K in terms of the Celsius temperature C. Transform the formula to find Celsius temperature in terms of Kelvin temperature. Then, find the Celsius temperature when the Kelvin temperature is 400°.

Solve for C.

K = C + 273.15

K – 273.15 = C + 273.15 – 273.15 Subtract 273.15 from each side.

K – 273.15 = C Simplify.

Find C when K = 400. 400 – 273.15 = C Substitute 400 for K.

126.85 = C

400º K is 126.85º C.

FormulasALGEBRA 1 LESSON 2-6

2-6

Solve each equation for the given variable.

1. 3x + 2y = z; y 2. = ; a

3. 3x + 4 = 2(3 – y); y

4. The formula v2 = can be used to find the constant speed v

of a satellite revolving around Earth in a circular orbit of radius r. G is

a number known as the gravitational constant and M is the mass of

Earth. Transform this formula to find the mass of Earth.

a – bc

d3

GMr

M = v2rG

FormulasALGEBRA 1 LESSON 2-6

2-6

a = + bcd3y = z – x1

232

y = – x + 132