34A Midterm 2 Review

Graph of 10^x

100

10^.4

2.5

200

log(2)

.3

300

log(700)

2.85

400

10^-1.4

.04

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10^-1.4=10^-2x10^.6=.01x4=.04

500

log(500³)

8.1

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log(500³)=3log(500)=3[log(100)+log(5)]= 3[2+.7]=6+2.1=8.1

Equations with Exponents

100

Solve for x:

10^x=100

x = 2

200

Solve for x:

2^x=8²

x = 6

----------

2^x=(2³)²=2⁶

300

Solve for x:

2(3^x)=4

x = log(2)/log(3)

----------

3^x=2

xlog(3)=log(2)

400

Solve for x:

3^x=9^x+1

x = -2

----------

3^x = (3²)^x+1=3^2(x+1)

x = 2x+2

500

Solve for x:

4^x=2(5^x-1)

x=log(2/5)/log(4/5)=[log(2)-log(5)]/[log(4)-log(5)]

----------

4^x=(5^x)(2/5)

xlog(4)=xlog(5)+log(2/5)

x[log(4)-log(5)]=log(2/5)

x = log(2/5)/[log(4)-log(5)]= log(2/5)/log(4/5)

Lines

100

Find the line that has slope 3 and goes through (0,1)

y = 3x+1

200

Find the line with slope -2 that goes through (3,5)

y = -2x+11

300

Find the line that passes through (2,2) and (4,6)

y = 2x-2

400

Find the intersection of y = 3x+1 and y = 2x+1

(0,1)

----------

3x+1=2x+1

x=0

500

Find the intersection of y = 2x+3 and the line with slope 1 that passes through (3,4).

(-2,-1)

----------

y = x+1

2x+3 = x+1

x = -2

Geometry

100

Express the area (A) of a square in terms of its perimeter (P).

A = (P/4)²

200

Express the perimeter (P) of the circle (fig. 1) in terms of the side length (s) of the square.

P = πs

300

Find the area (A) of the region outside of the circle (fig. 1) in terms of the radius (r) of the circle.

A=(4-π)r²

----------

Area of square = (side)²=(2r)²=4r²

Area of circle = πr²

Shaded area = (area of square)-(area of circle) = 4r²-πr²

400

Figure 2 shows a field. The fencing for it costs $10/m. How much does the fence cost?

(only the solid lines represent fences, not the dashed line)

Cost = $(15π+150)

----------

Outside perimeter of half-circle = π(diameter)/2 = 1.5π

Perimeter = 2(6)+3+1.5π = 15+1.5π

Cost = ($10)x(perimeter) = ($10)x(15+1.5π) =$(150+15π)

500

You have a box with a square base and no top with volume 400 cm³. The material for the base costs $3/cm²  and the material for the sides costs $2/cm². Express the cost of the box in terms of its height, h.

Cost = $1200/h+160h^1/2

----------

V = 400 cm³ = (area of base) x (height)

400/h = area of base = side², so side = 20/h^1/2

Area of box side = (height) x (side) = 20h^1/2

($3)x(area of base)+4x($2)x(area of box side)= $ (1200/h+160h^1/2)

Mixture Problems

100

You’re mixing 5L of a 20% bleach solution with 10L of a 10% bleach solution. How many liters of bleach are in the result?

                                   

    

2L

----------

.2(5)+.1(10)=1+1=2

200

You mix 3L of a 10% salt solution with 1L of a 18% salt solution. What's the percentage of salt in the mixture?

12% salt

----------

amount of salt in mix = .1(3)+.18(1) = .48

amount of mixture = 3+1 = 4

percentage of salt = (.48/4)x100% = 12%

300

You have 3L a mixture of 20% salt. How much water should you add to get a mixture of 10% salt?

3L

----------

x = amount of water

.2(3)+0(x)=.1(3+x)

.6=.1(3+x)

x = 3

400

You need 1 gallon of 30% alcohol solution, you have a 25% solution and a 40% solution. How much of the 25% solution should you use?

                                   

    

2/3 gallon

----------

x = gallons of 25%

.25x+.4(1-x)=.3(1)

-.15x=-.1

x=.1/.15=2/3

500

You have x mL of a solution that is 40% acid. You add a solution that is 10% acid until the result is only 30% acid. How much of the mixture do you end up with?

1.5x mL

----------

y = amount of mixture

.4x+.1(y-x) = .3y

.3x = .2y

y = .3x/.2 = 1.5x