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IrinaVladis [17]
3 years ago
8

Help me figure out how to simplify this please:

Mathematics
1 answer:
-BARSIC- [3]3 years ago
5 0
Think of it this way: 
8^3 means you have 3 times 8 multiplied together (8*8*8) 
Now this block is raised to the power of 7, so there are 7 blocks of 8*8*8 multiplied together. 
How many 8s do we have? 7 blocks of 3s, 7*3=21
Thus n=21
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What is the standard deviation of the data set given below?<br> 4,7, 8, 9, 12
larisa [96]

Answer:

I think 7

Step-by-step explanation:

4 , 8 , 9 and 12 has factors.

7 has no factor.

4= 2×2

8= 2×4

9= 3×3

12= 3×4

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3 years ago
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Evgen [1.6K]

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The answer is 8/5  or 8 out of 5.

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Which term describes the slope of the line below?
saul85 [17]

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negative

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Read 2 more answers
Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans
zloy xaker [14]
Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.



Part B:

The area of the square is given by

Area=(x+4)^2=x^2+8x+16

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

x^2+8x+16=6x+8 \\  \\ \Rightarrow x^2+8x-6x+16-8=0 \\  \\ \Rightarrow x^2+2x+8=0 \\  \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2}  \\  \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2}  \\  \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
7 0
3 years ago
Help pls I really need it
Talja [164]

Answer:

Infinite number of solutions

Step-by-step explanation:

when you solve for V, you notice that the term in "v" goes away, and you end up with a true statement:

2 = 2

This is a true statement no matter what values the variable V has, so it is true for all possible (infinite) values of "v".

8 0
3 years ago
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