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cricket20 [7]
3 years ago
12

Which calculation can be used to find the value of q in the equation q3 = 64?

Mathematics
2 answers:
alexandr402 [8]3 years ago
8 0
To isolate the variable q, divide both sides by 3:
q3 ÷ 3 = 64 ÷ 3
q = 21.33 or 21 1/3

If the q is cubed(I can't really tell), however, you would use the cube root:
³√q³ = ³√64
q = 4
Sauron [17]3 years ago
4 0
Cube root both q and 64

∛q³ = ∛64

q = ∛64

Simplify ∛64.

∛64 = ∛(4 x 4 x 4)

q = 4 is your answer

hope this helps
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The polygon is a regular nonagon (9-sides). What is the sum of the interior angles? What is the measurement for each angle?
lana66690 [7]
Divide the nonagon radially into 9 congruent, equilateral isosceles triangle. Each triangle has vertex angle of 360°/9 = 40°.
Interior angle of polygon = 180°-40° = 140°.
Sum of interior angles = 9(140°) = 1260°
6 0
3 years ago
What’s the inverse of 2(x-2)^2 = 8(7+y)
Anettt [7]

Answer:

y= (x-2)^2/4-7

Step-by-step explanation:

5 0
3 years ago
6f-7=29;5 what is this answer or how do you do i am having trouble finding it out
erik [133]
6(5)-7
30-7
23
The answer is false because it doest equal 29
8 0
3 years ago
Read 2 more answers
What is 1/2(4x+14)=2(x-7)
GREYUIT [131]

Answer:

no solution

Step-by-step explanation:

1/2(4x+14)=2(x-7)

Multiply both sides by 2.

4x + 14 = 4(x - 7)

4x + 14 = 4x - 28

Subtract 4x from both sides.

14 = -28

Since 14 = -28 is a false statement, there is no solution for this equation.

Answer: no solution

8 0
3 years ago
How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x​
Fofino [41]

Answer:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

Step-by-step explanation:

We are given the function:

f(x)=-4x^5-8x^3+12x

And we want to finds its zeros.

Therefore:

0=-4x^5-8x^3+12x

Firstly, we can divide everything by -4:

0=x^5+2x^3-3x

Factor out an x:

0=x(x^4+2x^2-3)

This is in quadratic form. For simplicity, we can let:

u=x^2

Then by substitution:

0=x(u^2+2u-3)

Factor:

0=x(u+3)(u-1)

Substitute back:

0=x(x^2+3)(x^2-1)

By the Zero Product Property:

x=0\text{ and } x^2+3=0\text{ and } x^2-1=0

Solving for each case:

x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}

Therefore, our real and complex zeros are:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

4 0
3 years ago
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