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:
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
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?
**SHOW YOUR WORK**
Thanks a lot.
- KarinLv 61 decade agoFavorite Answer
= 8 + 10
= 2(4x^2+4x+1) -3(2x+1) +5
f(2x) - 2g(x)
= (4x+10+3) -(6x-6-6)
grabage expected from a city of
= 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