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3 years ago

PLEASE HELP MEEEEEEEEEEEEE

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

5 0

Okay, so you have the last option (g(x)=3^x+2-7) as your first line at that is aligned with the x-axis then movies up slightly into Quad. one. Your answer for this is going to be... C. g(x)=3^x-2-7.

You are going to graph these by using only a few selected points, then finding your answer.

Your horizontal Asymptote is y=-7.

x--------------y

-2 -6.988

-1 -6.963

0 -6.889

2 -6

3 -4

Hope this helps you!!!

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Using the Factor Theorem, which of the polynomial functions has the zeros 2, radical 3 , and negative radical 3 ? f (x) = x3 – 2

butalik [34]

To solve this question, we use the factor theorem, and using it, the polynomial function is:

$f(x) = x^3 - 2x^2 - 3x + 6$

------------------------------

The factor theorem means that if k is a root of f(x), f(k) = 0.

Thus, applying the factor theorem for this question, we have to choose the function for which: $f(2) = 0, f(\sqrt{3}) = 0, f(-\sqrt{3}) = 0$

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Function 1:

$f(x) = x^3 - 2x^2 - 3x + 6$

Testing the values:

$f(2) = 2^3 - 2(2)^2 - 3(2) + 6 = 8 - 8 - 6 + 6 = 0$

$f(\sqrt{3}) = \sqrt{3^3} - 2(\sqrt{3})^2 - 3\sqrt{3} + 6 = \sqrt{3^2\times3} - 6 - 3\sqrt{3} + 6 = 3\sqrt{3} - 3\sqrt{3} = 0$

$f(-\sqrt{3}) = -\sqrt{3^3} - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 = -\sqrt{3^2\times3} - 6 + 3\sqrt{3} + 6 = -3\sqrt{3} + 3\sqrt{3} = 0$

Thus, since all three conditions are satisfied, $f(x) = x^3 - 2x^2 - 3x + 6$ is the polynomial function.

A similar question is given at brainly.com/question/11378552

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3 years ago

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Svetach [21]

Answer:

v = 25.6

Step-by-step explanation:

1. Multiply both sides of the equation by 4.

(6.4)(4) = 25.6

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Step-by-step explanation:

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