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Wewaii [24]
3 years ago
7

Which is the slope of the line that passes through the points (-2, 4) and (5, -1)?

Mathematics
2 answers:
Anton [14]3 years ago
8 0

<u>Answer:</u>

The slope of the line passing through the points (-2, 4) and (5, -1) is \frac{-5}{7}

<u>Solution:</u>

We know the slope intercept of the line is given by y = mx + c, where m is the slope of the line and c is a constant

To find the value of ‘m’ we use the given formula

m = \frac{(y_2- y_1)}{(x_2- x_1 )}

Here ,

x_1= -2

x_2= 5

y_1= 4

y_2= -1

Substituting the values in formula for m, we get

m=\frac{-1-4}{5-(-2)}

m=\frac{-1-4}{5+2)}

m=\frac{-5}{7)}

Therefore the slope of the line passing through the points (-2, 4) and (5, -1) is \frac{-5}{7}

coldgirl [10]3 years ago
4 0

Answer:

<h2>The slope is -5/7</h2>

Step-by-step explanation:

you can solve it many ways but i solved it this way

you take ur y2 and y1 and set it over x2 and x1

you get -1-4 / 5- -2

simply it

and it gives you -5/7

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In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank
fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) Amount \ of \, water \ remaining \ in \, the \ tank \ is \  \frac{x^3(6-\pi) }{6}

Step-by-step explanation:

1) Here we have;

First tank A

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The  volume of the sphere = \frac{4}{3} \pi r^3

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = \frac{4}{3} \pi \frac{x^3}{8}= \frac{1}{6} \pi x^3

For tank B

Volume of tank = x³

The  volume of the spheres = 8 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = 8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}

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The  volume of the spheres = 64 \times \frac{4}{3} \pi r^3

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r = x/8 and the volume of the sphere is thus;

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Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The  volume of the spheres = 512 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = 512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}

Amount \ of \ water \, remaining \, in \, the \ tank =  \frac{x^3(6-\pi) }{6}.

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3 years ago
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IrinaK [193]
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3 years ago
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Answer:

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