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timama [110]
2 years ago
15

Third-degree, with zeros of -3,-1, and 2, and passes through the point (3,6).

Mathematics
1 answer:
Y_Kistochka [10]2 years ago
3 0

Answer:

Below.

Step-by-step explanation:

In factor form this is:

f(x) = a(x + 3)(x + 1)(x - 2)     where a is a number to be found.

As it passes through (3, 6) we have:

6 = a(3 + 3)(3 + 1)(3 - 2)

24a = 6

a = 1/4.

So f(x) =  1/4(x + 3)(x + 1)(x - 2)

In general form this is:

f(x) = 1/4 x^3 + 1/2 x^2 - 5/4 x - 3/2

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3 years ago
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HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH
8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

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First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

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We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}

Thank you

8 0
2 years ago
five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. what is the solution
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