Well I'm just doing some random math so it might be 300?
Answer:
![\displaystyle Range: Set-Builder\:Notation → [y|-2 ≤ y] \\ Interval\:Notation → [-2, ∞) \\ \\ Domain: Set-Builder\:Notation → [x|-4 ≤ x] \\ Interval\:Notation → [-4, ∞)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Range%3A%20Set-Builder%5C%3ANotation%20%E2%86%92%20%5By%7C-2%20%E2%89%A4%20y%5D%20%5C%5C%20Interval%5C%3ANotation%20%E2%86%92%20%5B-2%2C%20%E2%88%9E%29%20%5C%5C%20%5C%5C%20Domain%3A%20Set-Builder%5C%3ANotation%20%E2%86%92%20%5Bx%7C-4%20%E2%89%A4%20x%5D%20%5C%5C%20Interval%5C%3ANotation%20%E2%86%92%20%5B-4%2C%20%E2%88%9E%29)
Explanation:
<em>See above graph</em>
I am joyous to assist you anytime.
Answer:
- Perimeter = 22*sqrt(2)
- Area = 60.5 inches
- D
Step-by-step explanation:
Remark
You need 2 facts.
- A square has 4 equal sides.
- It contains (by definition) 1 right angle but since we are not including and statement about parallel sides, it needs 4 right angles.
That means you can use the Pythagorean Theorem.
If one side of a square is a then the 1 after it is a as well.
Formula
- a^2 + a^2 = c^2
- 2a^2 = c^2
Givens
Solution
- 2a^2 = 11^2
- 2a^2 = 121 Divide by 2
- a^2 = 121/2 Take the square root of both sides
- sqrt(a^2) = sqr(121/2)
- a = 11/sqrt(2) Rationalize the denominator
- a = 11 * sqrt(2)/[sqrt(2) * sqrt(2)]
- a = 11 * sqrt(2) / 2
<em><u>Perimeter</u></em>
P = 4s
- P = 4*11*sqrt(2)/2
- P = 44*sqrt(2)/2
- P = 22*sqrt(2)
You don't need the area. The answer is D
<em><u>Area</u></em>
- Area = s^2
- Area = (11*sqrt(2)/2 ) ^2
- Area = 121 * 2 / 4
- Area = 60.5
Answer:
???69
Step-by-step explanation:
