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little asked in 科學及數學數學 · 1 decade ago

Math(Relations and Functions)急!!!15點

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



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


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.

1 Answer

  • Karin
    Lv 6
    1 decade ago
    Favorite Answer



    = 8 + 10

    = 18


    a. f(4)

    = -16-12+2


    b. f(-2m)




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

    = 8x^2+2x+4


    f(2x) - 2g(x)

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

    = -2x+25


    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


    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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