Answer:
A. reflection across the y-axiss
Step-by-step explanation:
Given:
The locations of the two points are (-4 , 8) and (-4 , -8).
To find:
The relation between two points.
From the given points (-4, 8) and (-4 , -8), it is clear that the y-coordinates are same but the sign of x-coordinates are opposite.
If a figure is reflected across the y-axis, then we change the sign of x-coordinate and the y-coordinates remain same, i.e.,
→
For (-4,8)
→ 
So, it is reflection across the y-axis.
Therefore, the correct option is A.
2304π is the answer and if you don’t want pi it’s 7238.2294738709 feet
Answer:
x = -1
Step-by-step explanation:
By graphing x = -1, you will see that the line splits this rhombus in two equal halves
Answer:

General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
- Coordinates (x, y)
- Slope Formula:

Step-by-step explanation:
<u>Step 1: Define</u>
Point (0, -5)
Point (-2, 1)
<u>Step 2: Find slope </u><em><u>m</u></em>
Simply plug in the 2 coordinates into the slope formula to find slope <em>m</em>
- Substitute in points [Slope Formula]:

- [Fraction - Denominator] Simplify:

- [Fraction - Numerator] Subtract:

- [Fraction - Denominator] Add:

- [Fraction] Divide:

we have a maximum at t = 0, where the maximum is y = 30.
We have a minimum at t = -1 and t = 1, where the minimum is y = 20.
<h3>
How to find the maximums and minimums?</h3>
These are given by the zeros of the first derivation.
In this case, the function is:
w(t) = 10t^4 - 20t^2 + 30.
The first derivation is:
w'(t) = 4*10t^3 - 2*20t
w'(t) = 40t^3 - 40t
The zeros are:
0 = 40t^3 - 40t
We can rewrite this as:
0 = t*(40t^2 - 40)
So one zero is at t = 0, the other two are given by:
0 = 40t^2 - 40
40/40 = t^2
±√1 = ±1 = t
So we have 3 roots:
t = -1, 0, 1
We can just evaluate the function in these 3 values to see which ones are maximums and minimums.
w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20
w(0) = 10*0^4 - 20*0^2 + 30 = 30
w(1) = 10*(1)^4 - 20*(1)^2 + 30 = 20
So we have a maximum at x = 0, where the maximum is y = 30.
We have a minimum at x = -1 and x = 1, where the minimum is y = 20.
If you want to learn more about maximization, you can read:
brainly.com/question/19819849