Answer:
1) 25
2) 2
3) f(g(1)) = 42
Step-by-step explanation:
1) Given that f(x) = 4x^2 + 9
If x = -2
f(-2) = 4(-2)^2 + 9
f(-2) = 4(4) + 9
f(-2) = 16 + 9
f(-2) = 25
2) Given that f(x) = 4x - 6
y = 4x - 6
Replace y with x
x = 4y - 6
MAke y the subject of the forfmula
4y = x+ 6
y = (x+6)/4
SInce x = 2
f^(-1)(2) = (2+6)/4
f^(-1)(2) = 8/4 = 2
3) If f(x) = 6x and g(x) = x+6
f(g(x)) = f(x+6)
f(x+6) = 6(x+6)
Since x = 1
f(g(1)) = 6(1+6)
f(g(1)) = 6(7)
f(g(1)) = 42
Answer:
Step-by-step explanation:
Answer:
{f(x) | -∞ < f(x) ≤ 4 }
Step-by-step explanation:
The range is made up of the y values. Remember f(x) = y
In this graph the least value of y is -∞ and the greatest value of y is 4
So, the range is {f(x) | -∞ < f(x) ≤ 4 } which is the 2nd choice
The slope-intercept form:
m - slope
b - y-intercept
We have the slope m = 5 → 
and point (-2, 0). Substitute:
<em>add 10 to both sides</em>
<h3>Answer: y = 5x + 10</h3>
Answer:
The initial speed of the car was 80 ft/s.
Step-by-step explanation:
The deceleration is the rate at which the car speed decreases. In this case the speed of the car goes all the way down to 0 ft/s and in order to do that it travelled 50 ft. So we will call the initial speed at which the car started to brake "v_0" and use Torricelli's equation to find it. The equation is given by:
v^2 = (v_0)^2 + 2*a*S
Where v is the final speed, v_0 is the initial speed, a is the rate of acceleration and S is the space travelled. Using the values that the problem gave to us we have:
0^2 = (v_0)^2 - 2*64*50
0 = (v_0)^2 - 6400
(v_0)^2 = 6400
v_0 = sqrt(6400) = 80 ft/s
Notice that in this case "a" was negative, since the car was decelerating instead of accelerating.
The initial speed of the car was 80 ft/s.
