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3 years ago

Need help with this math question. I have tried my best to solve it but I dont really know.

Mathematics

2 answers:

MissTica3 years ago

8 0

Answer:

all real numbers

Step-by-step explanation:

I cannot exactly explain it since you could geta. multitude of different answers for the problem

Alika [10]3 years ago

8 0

Answer:

The answer to your question is: No solution

Step-by-step explanation:

-2(6 + s) ≥ - 15 - 2s

-12 - 2s ≥ -15 - 2s

-2s + 2s ≥ -15 + 12

0 ≥ -3

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Given the graph of f ′(x), the derivative of f(x), which of the following statements are true about the graph of f(x)?

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4 years ago

Question

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 p and q are the zeros.

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

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

Therefore, we acquire:

$f(x)=a(x-(2))(x-(-8i))(x-(8i))$

Simplify:

$f(x)=a(x-2)(x+8i)(x-8i)$

Expand the second and third factors:

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

Hence, our function is now:

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

It equals 320 when x = 4. Therefore:

$320=a(4-2)(4^2+64)$

Solve for a. Evaluate:

$320=(2)(80)a$

So:

$320=160a\Rightarrow a=2$

Our third-degree polynomial equation is:

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

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3 years ago