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V125BC [204]
3 years ago
8

Simplify the radical expression. square root of 39 (square root of 6 + 7)

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
7 0
The solution to the problem is as follows:

If you meant, Square root of 32: 
<span>
32 </span>
^ 
16 2 
^ 
8 2 
^ 
4 2 
^ 
2 2 

<span>So the answer would be 4 root 2 </span>

<span>So Im not sure for 39 </span>

39 
^ 
<span>3 ...
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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12 out of 30 tickets sold were early admission what percentage of tickets were early admission tickets
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Step-by-step explanation:

12 divided by 30 will give you 0.4

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16 factory workers made 40 pieces in a certain amount of time. How many workers are needed to make (1) 20, (2)25 and (3)100 piec
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Read 2 more answers
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}\}

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3 years ago
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