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Alex_Xolod [135]
3 years ago
13

Find the value of y please ASAP

Mathematics
2 answers:
nekit [7.7K]3 years ago
5 0

x/3 = 12/x

x^2 = 36

x = 6

9/y = y /x

9/y = y / 12

y^2 = 9 * 12

y = 9 * 4 * 3

y = 6√3

Answer

y = 6√3

konstantin123 [22]3 years ago
4 0

Right Triangle Altitude Theorem

Draw the altitude of a right triangle from the vertex of the right angle to the hypotenuse, then:

Part a: the measure of the altitude is the geometric mean between the measures of the two segments of the hypotenuse.

Part b: each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg.

Using Part b,

\dfrac{12}{y} = \dfrac{y}{9}

y^2 = 12 \times 9

y = \sqrt{9 \times 4 \times 3}

y = 3 \times 2 \times \sqrt{3}

y = 6\sqrt{3}

Answer: ~~~6\sqrt{3}

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3 years ago
What are the solutions
netineya [11]

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A volleyball reaches its maximum height of 13 feet, 3 seconds after its served. Which of the following quadratics could model th
julsineya [31]
Only selections B and D give a maximum height of 13 at t=3. However, both of those functions have the height be -5 at t=0, meaning the ball was served from 5 ft below ground. This does not seem like an appropriate model.

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8 0
3 years ago
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7=14x-42y solve equation for y
dedylja [7]
7 = 14x - 42y
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3 0
3 years ago
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The path followed by a roller coaster as it climbs up and descends down from a peak can be modeled by a quadratic function, wher
Romashka [77]

Answer:

y = -0.1x^2 + 250 ft

Step-by-step explanation:

Because this quadratic equation would have the curve-down form of:

y = -ax^2 + b

where a and b are positive coefficient.

If we let the peak (250 ft) of the curve be at x = 0. Then

y = -a0^2 + b = 250

b = 250

Also at the begins and ends, thats where y = 0, the 2 points are separated by 100 ft. So let the begin at -50 ft and the end at 50ft. We have

-a(\pm 50)^2 + 250 = 0

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a = 250/2500 = 0.1

Therefore, the model quadratic equation of our path would be

y = -0.1x^2 + 250 ft

8 0
3 years ago
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