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3 years ago

7

Help.help.help.pls.pls

2 answers:

ycow [4]3 years ago

7 0

<h3>Answer: 3</h3>

========================================

Work Shown:

Solve the equation for y

-13y - 5 = 8

-13y = 8+5

-13y = 13

y = 13/(-13)

y = -1

Then use this y value to compute 4y+7

4y+7 = 4(-1)+7 = -4+7 = 3

timofeeve [1]3 years ago

5 0

-13y-5=8

Take 5 to the other side as a positive number

-13y=13

Divide both sides by -13

Therefore, y is equal to -1

But, the question isn't answered, we need to substitute the value of y into 4y+7, to fully answer the question

With our knowledge that y is equal to -1, the expression becomes

4*-1+7, which is 3

So, your answer is 3

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4 years ago

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Kay [80]

Answer:

$f(x)=2(x-2)(x^2+64)$

Step-by-step explanation:

A standard polynomial in factored form is given by:

$f(x)=a(x-p)(x-q)...$

Where <em>p</em> and <em>q</em> are the zeros.

We want to find a third-degree polynomial with zeros <em>x</em> = 2 and <em>x </em>= -8i and equals 320 when <em>x </em>= 4.

First, by the Complex Root Theorem, if <em>x</em> = -8i is a root, then <em>x </em>= 8i must also be a root.

Therefore, we acquire:

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$=(x+8i)x+(x+8i)(-8i)\\\\=(x^2+8ix)+(-8ix-64i^2)\\\\=(x^2)+(8ix-8ix)+(-64i^2)\\\\=x^2-64(-1)\\\\ =x^2+64$

Hence, our function is now:

$f(x)=a(x-2)(x^2+64)$

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$320=160a\Rightarrow a=2$

Our third-degree polynomial equation is:

$f(x)=2(x-2)(x^2+64)$

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3 years ago