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2 years ago

Solve for x. 6x + 9 + 4x + 2 = 21

Mathematics

2 answers:

Pie2 years ago

5 0

Answer:

$6x + 9 + 4x + 2 = 21 \\ 6x + 4x + 11 = 21 \\ 10x + 11 = 21 \\ 10x = 21 - 11 \\ 10x = 10 \\ x = \frac{10}{10} \\ x = 1$

<h3> <u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u></h3>

Mama L [17]2 years ago

5 0

Answer:

i got u!!!! it's x=1

Step-by-step explanation:

sooo basically isolate the variable by dividing each side by factors that don't contain the variable!

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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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2 years ago

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agasfer [191]

6x - 2x = 16

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