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Ksivusya [100]
3 years ago
9

Simplify 4√32 A. 4√8 B. 16√2 C. 8√2 D. 24

Mathematics
1 answer:
denpristay [2]3 years ago
4 0
4\sqrt{32} =\\\\ 4\sqrt{16\times 2} =\\\\4\sqrt{16}\times \sqrt{2} =\\\\ 4\times 4\times \sqrt{2} =\\\\\ 16\times \sqrt{2}=\\\\ 16\sqrt{2}

Your final answer is B.
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A regular octagon has side lengths of 8 centimeters. What is the approximate area of the octagon?
lara31 [8.8K]

Answer:

The answer is B if the octagon has side lengths of 8

5 0
2 years ago
What is the system of solution to y-x=-13 and -4x+3y=-51
poizon [28]
Equation 1 ==> y - x = -13
Equation 2 ==> -4x + 3y = -51
3(y - x) = 3(-13)
Equation 3 ==> 3y - 3x = -39

Equation 2 - 3
= (3y - 3y) + ( -4x - (-3x) ) = -51 - (-39)
-x = -12
x = 12

Substitude x into equation 1
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8 0
4 years ago
if you are 15 years old now, which equation could be used to solve for your age, y, 12 years from now ?
Vaselesa [24]

Answer:

Hi there!

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3 0
3 years ago
Use the Quadratic formula to find all real zeros of the 2nd degree polynomial
schepotkina [342]

Answer:

  x ∈ {-5, -1}

Step-by-step explanation:

Here's the solution using the quadratic formula:

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-24\pm\sqrt{24^2-4\cdot 4\cdot 20}}{2\cdot 4}\\\\=\dfrac{-24\pm\sqrt{576-320}}{8}=\dfrac{-24\pm\sqrt{256}}{8}\\\\=\dfrac{-24\pm 16}{8}=-3\pm 2=\{-5,-1\}

The real zeros are -5 and -1.

_____

There are many ways to check your answer. One of them is to look at the given quadratic, which has no changes of sign in its coefficients. (They are all positive.) That means there can be no positive real roots, so already you know that x=0.5 won't work.

Also, the constant in the quadratic is the product of the roots, For your roots, their product is -7/4, so even multiplying by 4 (the leading coefficient in the given quadratic), you don't get anything like 20.

4 0
3 years ago
N to the second power - 5 in - 1 in equals 6​
Wittaler [7]

Answer:

the question is incorrect.

4 0
3 years ago
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