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3 years ago

Select the quadratic that has roots x=8 and x=-5 a x^2 + 3x -40

Mathematics

1 answer:

sergiy2304 [10]3 years ago

4 0

If x = 8, then the factor is (x - 8); likewise, if x = -5, then the factor is (x + 5). FOIL those together and get B

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The coordinates of quadrilateral PQRS are P(–3, 0), Q(0, 4), R(4, 1), and S(1, –3). What best describes the quadrilateral?

denis-greek [22]

Answer:

Square

Step-by-step explanation:

Plot the vertices of the quadrilateral PQRS on the coordinate plane (see attached diagram). The diagram shows that this is a square. Let's prove it.

1. Find all sides lengths:

$PQ=\sqrt{(-3-0)^2+(0-4)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \\QR=\sqrt{(0-4)^2+(4-1)^2}=\sqrt{(-4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}=5\\ \\RS=\sqrt{(4-1)^2+(1-(-3))^2}=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \\SP=\sqrt{(1-(-3))^2+(-3-0)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$

All sides have the same lengths.

2. Find the slopes of all lines:

$PQ:\ \dfrac{4-0}{0-(-3)}=\dfrac{4}{3}\\ \\QR:\ \dfrac{1-4}{4-0}=-\dfrac{3}{4}\\ \\RS:\ \dfrac{-3-1}{1-4}=\dfrac{4}{3}\\ \\SP:\ \dfrac{0-(-3)}{-3-1}=-\dfrac{3}{4}$

Since the slopes of PQ and RS are the same, lines PQ and RS are parallel. Since the slopes of QR and SP are the same, lines QR and SP are parallel.

The slopes $\frac{4}{3}$ and $-\frac{3}{4}$ have the product of

$-\dfrac{3}{4}\cdot \dfrac{4}{3}=-1,$

then lines are pairwise perpendicular.

This means PQRS is a square.

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Consider the procedure used below to solve the given equation.

muminat

The first mistake was made in step 1 because they took out the x’s it should be 10x-25+2x+8=43

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Answer:

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Sorry if this is the wrong answer. But I hope you get a good grade on your test.

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