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OLEGan [10]
3 years ago
15

Please help me this is for my exam and i need to pass this or i won’t graduate. please !!!!!!!!

Mathematics
1 answer:
kakasveta [241]3 years ago
3 0

Answer:

C. -3 must be a root of the polynomial 2·x² + 9·x + 9

Step-by-step explanation:

Given that the remainder when (2·x² + 9·x + 9) is divided by (x + 3) equals zero, then (x + 3) is a factor of 2·x² + 9·x + 9 and -3 is a root of the polynomial

Explaining by using the long division of the polynomial is presented as follows;

<u>2·x + 3</u>

(2·x² + 9·x + 9)/(x + 3)

<u>- (2·x² + 6·x)</u>

3·x + 9

-(3·x + 9)

0

Therefore, (2·x + 3) × (x + 3) = 2·x² + 9·x + 9

The solution of the 2·x² + 9·x + 9 = 0 gives the root of the polynomial

The root is given by, 2·x² + 9·x + 9 = (2·x + 3) × (x + 3) = 0

Therefore, the roots are;

2·x = -3 or x = -3/2 and x = -3

Therefore;

-3 must be a root of the polynomial 2·x² + 9·x + 9.

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Answer:

\sf R=\left(3, -\dfrac{5}{4}\right)

Step-by-step explanation:

Given:

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  • Q = (-2, 3)
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<u />

<u>Distance between two points</u>

\sf d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

(where (x₁, y₁) and (x₂, y₂) are the two points)

Use the <u>distance formula</u> to derive equations for PR and QR.

Let (x₁, y₁) = P = (1, 5)

Let (x₂, y₂) = R = (3, b)

\begin{aligned} \sf PR & =\sf \sqrt{(3-1)^2+(b-5)^2}\\ & = \sf \sqrt{4+(b-5)^2} \end{aligned}

Let (x₁, y₁) = Q = (-2, 3)

Let (x₂, y₂) = R = (3, b)

\begin{aligned} \sf QR & =\sf \sqrt{(3-(-2))^2+(b-3)^2}\\ & = \sf \sqrt{25+(b-3)^2} \end{aligned}

As PR = QR, equate the derived equations and solve for b:

\begin{aligned} \sf PR & = \sf QR \\\sf \sqrt{4+(b-5)^2} & = \sf \sqrt{25+(b-3)^2}\\\sf 4+(b-5)^2 & = \sf 25+(b-3)^2\\\sf 4+b^2-10b+25 & = \sf 25 + b^2-6b+9\\\sf b^2-10b+29 & = \sf b^2 -6b +34\\\sf -10b+29 & = \sf -6b + 34\\\sf -4b & = \sf 5\\\sf b & = -\dfrac{5}{4}\end{aligned}

Substitute the found values of a and b to find the coordinates of R:

\sf R=(a,b)=\left(3, -\dfrac{5}{4}\right)

Learn more about the distance formula here:

brainly.com/question/28144723

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