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Andrews [41]
3 years ago
11

(-5, -11) and (-2, 1)

Mathematics
2 answers:
irinina [24]3 years ago
8 0

Answer:

the answer is 0 because it 3 negative number

Stolb23 [73]3 years ago
5 0
The answer is 0 because they are negatives
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ACT scores have a mean of 21.5 and 26 percent of the scores are above 24. The scores have a distribution that is approximately n
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A photographer is 6 feet tall and cast a shadow that is 2 feet long. He is photographing a palm tree with a shadow that is 7 fee
Novay_Z [31]

Answer: The height of the tree is 21 feet.

Step-by-step explanation:

Here we can assume that the angle at which the sun impacts the photographer and the tree to be the same angle.

Then we can think in both cases as triangles rectangles, where the height is a cathetus, and the shadow is the other cathetus.

Then we will have a relationship like:

Tg(angle) = shadow/height

height = shadow/Tg(angle)

Now, because for both triangles we have the same angle, then Tg(angle) will be the same number for both cases, and we can just think of it as constant K

Tg(angle) = K

Then we have the equation:

Height = Shadow/K

We know that the photographer is 6ft tall, and his shadow is 2 ft long, we can replace those two things in the above equation and find the value of k:

6ft = 2ft/K

K = 2ft/6ft = (1/3)

Now we know that the shadow of the tree is 7ft long, then the height will be:

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What’s the diameter of a circle 5in
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A household aquarium tank in the shape of a rectangular prism has a base length of 242424 inches (\text{in})(in)(, start text, i
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Given

l =24in --- length

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Required

The absolute change in the height of the water

First, calculate the base area (b):

b = l * w

b =24in * 15in

b =360in^2

The height of the water that was removed is:

<em />H = \frac{V_1}{b}<em> i.e. the volume of the water removed divided by the base area</em>

H = \frac{900in^3}{360in^2}

H = \frac{900in}{360}

H = 2.5in

The absolute change in height is:

\triangle H = |h - H|

\triangle H = |12in - 2.5in|

\triangle H = |9.5in|

\triangle H = 9.5in

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