Let the distance of Judy from intersection is x and distance of Jackie from intersection is y.
We convert the given information to equations, step by step.
Point 1: When Jackie is 1 mile farther from the intersection than Judy.
This means y is 1 mile more than x.
So,
y = 1 + x
Point 2: The distance between them is 2 miles more than Judy’s distance from the intersection.
Distance between is x+ y.
So, x+y is 2 miles more than y.
x+y = y + 2
⇒
x = 2
From point 1 we have:
y = 1 + x = 1+ 2 = 3
So,
Distance of Judy from intersection is 2 miles and distance of Jackie from intersection is 3 miles.
Answer:
49°
Step-by-step explanation:
arccos x = 15/23
Answer:
Check step by step explanation
Step-by-step explanation:
a) Let x represent the amount of rides she takes, the 6 and 2.5 represent the amount of money required to enter
A(x) = 1.5x + 6
B(x) = 2x + 2.5
b) You can find this by making the 2 equations equal to each other and solving for x
1.5x + 6 = 2x + 2.5
3.5 = .5x
7 = x
7 rides make them equal cost
c) Plug in 5 for each of the equations and find out which one is cheaper
A(5) = 1.5 * 5 + 6
A(5) = $13.50
B(5) = 2 * 5 + 2.5
B(5) = $12.50
Carnival B is cheaper
ANSWER
1. k=13
2. x=-10
EXPLANATION
The given function is
![f(x) = {x}^{2} + 3x - 10](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%20%20%2B%203x%20%20-%2010)
To find f(x+5), plug in (x+5) wherever you see x.
This implies that:
![f(x) = {(x + 5)}^{2} + 3(x + 5) - 10](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%7B%28x%20%2B%205%29%7D%5E%7B2%7D%20%20%20%2B%203%28x%20%2B%205%29%20-%2010)
Expand:
![f(x) = {x}^{2} + 10x + 25+ 3x + 15- 10](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%20%7Bx%7D%5E%7B2%7D%20%20%20%20%2B%2010x%20%2B%2025%2B%203x%20%2B%2015-%2010)
Simplify to obtain
![f(x) = {x}^{2} + 13x + 30](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%20%7Bx%7D%5E%7B2%7D%20%20%20%20%2B%2013x%20%2B%2030)
We now compare with,
![f(x) = {x}^{2} + kx + 30](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%20%7Bx%7D%5E%7B2%7D%20%20%20%20%2B%20kx%20%2B%2030)
This implies that:
![k = 13](https://tex.z-dn.net/?f=k%20%3D%2013)
To find the smallest zero of f(x+5), we equate the function to zero and solve for x.
![{x}^{2} + 13x + 30 = 0](https://tex.z-dn.net/?f=%7Bx%7D%5E%7B2%7D%20%20%20%20%2B%2013x%20%2B%2030%20%3D%200)
![{x}^{2} + 10x + 3x + 30 = 0](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%20%2B%2010x%20%2B%203x%20%2B%2030%20%3D%200)
![x(x + 10) + 3(x + 10) = 0](https://tex.z-dn.net/?f=x%28x%20%2B%2010%29%20%2B%203%28x%20%2B%2010%29%20%3D%200)
![(x + 3)(x + 10) = 0](https://tex.z-dn.net/?f=%28x%20%2B%203%29%28x%20%2B%2010%29%20%3D%200)
![x = - 10 \: or \: x = - 3](https://tex.z-dn.net/?f=x%20%3D%20%20-%2010%20%5C%3A%20or%20%5C%3A%20x%20%3D%20%20-%203)
The smallest zero is -10.