Answer:
Step-by-step explanation:
carnival A : $ 6 dollar admission and $ 1.50 per ride
A(x) = 1.5x + 6
carnival B : $ 2.50 admission and $ 2 per ride
B(x) = 2x + 2.5
how many rides can Marnie go on so that the total cost of attending both is the same...
1.5x + 6 = 2x + 2.5
6 - 2.5 = 2x - 1.5x
3.5 = 0.5x
3.5 / 0.5 = x
7 = x <=== she would have to go on 7 rides for them to both cost the same
Answer:
c) 3 units
d) g(x) - f(x) = x² + 2x
e) (-∞, -2] ∪ [0, ∞)
Step-by-step explanation:
<h3><u>Part (c)</u></h3>
To calculate the length of FC, first find the coordinates of point C.
The y-value of point C is zero since this is where the function f(x) intercepts the x-axis. Therefore, set f(x) to zero and solve for x:
![\implies 1-x^2=0](https://tex.z-dn.net/?f=%5Cimplies%201-x%5E2%3D0)
![\implies x^2=1](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2%3D1)
![\implies \sqrt{x^2}=\sqrt{1}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%7Bx%5E2%7D%3D%5Csqrt%7B1%7D)
![\implies x= \pm 1](https://tex.z-dn.net/?f=%5Cimplies%20x%3D%20%5Cpm%201)
As point C has a <u>positive</u> x-value, C = (1, 0).
To find point F, substitute the x-value of point C into g(x):
![\implies g(1)=2(1)+1=3](https://tex.z-dn.net/?f=%5Cimplies%20g%281%29%3D2%281%29%2B1%3D3)
⇒ F = (1, 3).
Length FC is the <u>difference</u> in the y-value of points C and F:
![\begin{aligned} \implies \sf FC& = \sf y_F-y_C\\ & = \sf 3-0\\ & =\sf 3\:units \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5Cimplies%20%5Csf%20FC%26%20%3D%20%5Csf%20y_F-y_C%5C%5C%20%26%20%3D%20%5Csf%203-0%5C%5C%20%26%20%3D%5Csf%203%5C%3Aunits%20%5Cend%7Baligned%7D)
<h3><u>Part (d)</u></h3>
Given functions:
![\begin{cases}f(x)=1-x^2\\ g(x)=2x+1 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Df%28x%29%3D1-x%5E2%5C%5C%20g%28x%29%3D2x%2B1%20%5Cend%7Bcases%7D)
Therefore:
![\begin{aligned}\implies g(x)-f(x) & = (2x+1) - (1-x^2)\\& = 2x+1-1+x^2\\& = x^2+2x\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20g%28x%29-f%28x%29%20%26%20%3D%20%282x%2B1%29%20-%20%281-x%5E2%29%5C%5C%26%20%3D%202x%2B1-1%2Bx%5E2%5C%5C%26%20%3D%20x%5E2%2B2x%5Cend%7Baligned%7D)
<h3><u>Part (e)</u></h3>
The values of x for which g(x) ≥ f(x) are where the line of g(x) is above the curve of f(x):
Point A is the <u>y-intercept</u> of both functions, therefore the x-value of point A is 0.
To find the x-value of point E, equate the two functions and solve for x:
![\begin{aligned}g(x) & = f(x)\\\implies 2x+1 & = 1-x^2\\x^2+2x & = 0\\x(x+2) & = 0\\\implies x & = 0, -2\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dg%28x%29%20%26%20%3D%20f%28x%29%5C%5C%5Cimplies%202x%2B1%20%26%20%3D%201-x%5E2%5C%5Cx%5E2%2B2x%20%26%20%3D%200%5C%5Cx%28x%2B2%29%20%26%20%3D%200%5C%5C%5Cimplies%20x%20%26%20%3D%200%2C%20-2%5Cend%7Baligned%7D)
As the x-value of point E is negative ⇒ x = -2.
Therefore, the values of x for which g(x) ≥ f(x) are:
- <u>Solution</u>: x ≤ -2 or x ≥ 0
- <u>Interval notation</u>: (-∞, -2] ∪ [0, ∞)
Answer:
The answer is 699...............
Answer:
B
Step-by-step explanation:
Product is what you get when you multiply