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Leviafan [203]
3 years ago
7

Please help asap! i need help :(

Mathematics
1 answer:
victus00 [196]3 years ago
4 0
C i just took the test hope this helps:-)
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Solve the equation 6x to the 3rd power = -1297 then round answer to two decimal places when appropriate
love history [14]
I got 213 . because 6x6x6. equal 213.
6 0
3 years ago
Find the total area the regular pyramid.<br><br><br><br> T. A. =
alexira [117]
The answer would be   A = 54raiz (3) + 18raiz (91)
Formula:
 A = Ab + Al Where, Ab=base area A= lateral area

 The area of the base is: Ab = (3/2) * (L ^ 2) * (root (3)) Where, L= side of the hexagon. Substitute: Ab = (3/2) * (6 ^ 2) * (root (3)) Ab = (3/2) * (36) * (root (3)) Ab = 54raiz (3)


 The lateral area is: Al = (6) * (1/2) * (b) * (h) Where, b= base of the triangle h= height of the triangle Substitute: Al = (6) * (1/2) * (6) * (root ((8) ^ 2 + ((root (3) / 2) * (6)) ^ 2)) Al = 18 * (root (64 + 27)) Al = 18raiz (91)

 The total area is: A = 54raiz (3) + 18raiz (91)

6 0
3 years ago
Read 2 more answers
Consider the compound interest equation B(t)=100(1.1664)t. Assume that n=2, and rewrite B(t) in the form B(t)=P(1+rn)nt. What is
Helen [10]
The answer is I think 108
3 0
3 years ago
The function f(t) represents the cost to connect to the Internet at an online gaming store. It is a function of t, the time
EleoNora [17]

The true statement about the Internet connection cost is D. Any amount of time over an hour and a half would cost $10.

<h3>How to explain the cost?</h3>

In this situation, the function is f (t), when t is a value between 0 and 30. The cost is $0 for the first 30 minutes

When t is a value between 30 and 90, the cost is US$ 5 if the connection takes between 30 and 90 minutes.

Here, the true statement about the Internet connection cost is D. Any amount of time over an hour and a half would cost $10.

Learn more about cost on:

brainly.com/question/13862342

#SPJ1

7 0
2 years ago
Can somebody please help me with this?
ser-zykov [4K]

Answer:

d fasho

Step-by-step explanation:

8 0
2 years ago
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