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3 years ago

How many area codes (ABC) would be possible if the first 2 digits can be any number 2-9 & last digit can be any number 0-9?

1 answer:

8 0

The answer would be A. 640

Possibilities for the first two are 8 and 8 (2-9 is 8 numbers) and the last is 10 (0-9 is 10 numbers)

So the probability is found by multiplying 8*8*10 which is 640

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If you have 8 batteries cost $40 [cost per battery]

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Answer:

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3 years ago

Read 2 more answers

Any volunteer please help

den301095 [7]

Answer:

Part A)

The equation in the point-slope form is:

$y-11=\frac{4}{3}\left(x-\left(-2\right)\right)$

Part B)

The graph of the equation is attached below.

Step-by-step explanation:

Part A)

Given

The point = (-2, 11)

m = 4/3

The point-slope form of the line equation is

$y-y_1=m\left(x-x_1\right)$

Here, m is the slope and (x₁, y₁) is the point

substituting the values m = 4/3 and the point (-2, 11) in the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

$y-11=\frac{4}{3}\left(x-\left(-2\right)\right)$

Thus, the equation in the point-slope form is:

$y-11=\frac{4}{3}\left(x-\left(-2\right)\right)$

Part B)

As we have determined the point-slope form which passes through the point (-2, 11) and has a slope m = 4/3

The graph of the equation is attached below.

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