# Math(Relations and Functions)急!!!15點

1. If two functions are definded as: f(x)=3x-1 and g(x)=x^2+1,

Calculate:

f(3)+g(-3)

2. If f(x)=-x^2-3x+2, evalute the values of:

a. f(4)

b. f(-2m)

3. If f(x)=2x^2-3x+5, evaluate the expression when f(2x+1)

4. Determine the expression for: f(2x)-2g(x) if we use the functions

f(x)=2(x+5)+3 and g(x)=3(x-1)-3

5. The amount of grabage collected in a city varies directly as the

population. If a city of 30000 people generates 132 tonnes of grabage,

how many tonnes of grabage would you expect from a city of

67000 people?

6. An object is supported from the end of a spring as shown in the

diagram to the right.

The distace the spring stretches varies partially to the mass of the

object.

When a 2kg mass is supported, the spring stretches 6cm.

When a 6kg mass is supported, the spring stretches 12cm.

a. What is the partial variation equation for the above situation.

b. Use the partial variation equation to determine how many cm a

4kg mass should stretch the spring.

7. The cost of renting a car for a day varies partially with the distance

driven. It costs \$54 to drive 400km and it costs \$72 to drive 600km.

a. What is the fixed cost?

b. What is the cost per km?

c. What is the partial varation equation?

(Let C=cost and k=km driven. Write your equation in form

C=(cost per km)k+fixed cost)

d. How far could a person drive for \$90?

Thanks a lot.

Rating

1.

f(3)+g(-3)

= 8 + 10

= 18

2.

a. f(4)

= -16-12+2

=-26

b. f(-2m)

=-4m^2+6m+2

3.

f(2x+1)

= 2(4x^2+4x+1) -3(2x+1) +5

= 8x^2+2x+4

4.

f(2x) - 2g(x)

= (4x+10+3) -(6x-6-6)

= -2x+25

5.

grabage expected from a city of

67000 people

= 67000/30000x132 tonnes

= 294.8 tonnes

6. a. partial variation equation for the above situation:

y = kx + h

6 = 2k+h and 12 = 6k+h

Therefore, k = (12-6)/4 = 1.5 and h = 6-3=3

i.e. y = 1.5x+3

b. 4kg mass should stretch the spring

= 6+3 = 9cm

7.

Partial variation equation : C = (cost per km)k + (fixed cost)

54 = (cost per km)400 + (fixed cost)

72 = (cost per km)600 + (fixed cost)

cost per km = (72-54)/200 = \$0.09

fixed cost = 54 - 0.09x400 = \$18

C = 0.09k +18

If C = \$90, k = (90-18)/0.09 = 800km

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