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3 years ago

-4(3x-1)=12-8x and plz show how you got it

Mathematics

2 answers:

ioda3 years ago

6 0

-12x + 4 = 12 - 8x
8x. 8x

-4x + 4 = 12
-4. -4

-4x = 8

x = -2 is the answer

Inga [223]3 years ago

3 0

Step 1: -4(3x-1) becomes -12x+4
step 2: -12x+4=12-8x
step 3: add 12x to both sides, 4=12+4x
step 4: subtract twelve from both sides -8=4x
step 5: divide by 4, x=-2

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2 years ago

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

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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.

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

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

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

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This means PQRS is a square.

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