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3 years ago

Find the slope of the line through each pair of points. (2, -4), (5,-8)

Mathematics

1 answer:

slega [8]3 years ago

8 0

Answer:

$m=-\frac{4}{3}$

Step-by-step explanation:

$Slope=m=\frac{y_2-y_1}{x_2-x_1}$

Here

$(x_1, y_1)=(2, -4) \ \ and \ \ (x_2, y_2)=(5, -8)$

$m=\frac{-8-(-4)}{5-2}$

$m=-\frac{4}{3}$

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3 years ago

How many different 10-letter words (real or imaginary) can be formed from the letters of the word LITERATURE?

nordsb [41]

Answer:

The number of words that can be formed from the word "LITERATURE" is 453600

Step-by-step explanation:

Given

Word: LITERATURE

Required: Number of 10 letter word that can be formed

The number of letters in the word "LITERATURE" is 10

But some letters are repeated; These letters are T, E and R.

Each of the letters are repeated twice (2 times)

i.e.

Number of T = 2

Number of E = 2

Number of R = 2

To calculate the number of words that can be formed, the total number of possible arrangements will be divided by arrangement of each repeated character. This is done as follows;

Number of words that can be formed = $\frac{10!}{2!2!2!}$

Number of words = $\frac{10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1}{2 * 1 * 2 * 1 * 2 * 1}$

Number of words = $\frac{3628800}{8}$

Number of words = 453600

Hence, the number of words that can be formed from the word "LITERATURE" is 453600

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3 years ago

If E F = 7 , A C = 16 , and A F = 9

Vilka [71]

Answers:

CB = 14

GF = 8

FB = 9

EF is parallel to CB

====================================

Explanations:

Points E and F are midpoints of their respective sides. They form the midsegment EF. Because EF is a midsegment, A midsegment is half the length of its parallel counterpart, so CB is two times longer than EF. If EF is 7 units long, then CB = 2*EF = 2*7 = 14

For similar reasons, GF is parallel to AC. If AC = 16, then half of that is GF = (1/2)*AC = 0.5*16 = 8.

FB = FA = 9 as these segments have the same single tickmark to indicate they are the same length

EF is parallel to CB because EF is a midsegment, and this is one of the properties of being a midsegment. We can show that quadrilateral EGBF is a parallelogram to help prove this.

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3 years ago