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Sedaia [141]
1 year ago
6

A rocket is launched from the top of a 70 foot cliff with an initial velocity of 130 feet per second. The height, h, of the rock

et after t seconds is given by the equation
h=-161² +130t+70. How long after the rocket is launched will it be 30 feet from the ground?
Mathematics
1 answer:
lana [24]1 year ago
6 0

Answer:

The answer is 7.8 seconds.

  1. Solve for t in the equation using the Quadratic formula.
  2. Eliminate the negative solution and the positive solution is your answer.

Please Mark Brainliest If This Helped!

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Someone who actually knows what they doing please ‍♀️
mariarad [96]

Answer:

5/9 (F-32) = 25

Step-by-step explanation:

5/9 (F-32) = C

Multiply each side by 9/5

5/9 * 9/5 (F-32) =9/5 C

F -32 = 9/5 C

Add 32

F -32 +32 = 9/5 C +32

F = 9/5 C +32

We want 25 C to Fahrenheit

F = 9/5 *25 +32

F = 9*5 +32

F = 45+32

F = 77

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3 years ago
Quadrilateral ABCD is dilated by a scale factor of 1/3 centered around (1,2). Which statement is true about the dilation
morpeh [17]

Answer:

I cannot help answer this question if you don't add the statements...

Step-by-step explanation:

5 0
3 years ago
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Choose the best answer:
Oksi-84 [34.3K]

Answer:

The length of the hypotenuse is 2 square root of 13 ⇒ c

Step-by-step explanation:

The rule of the area of the right triangle is A = \frac{1}{2} × leg1 × leg2, where

leg1 and leg2 are the sides of the right angle

∵ The area of a right triangle is 12 in²

∵ The ratio of the length of its legs is 2: 3

→ Let leg1 = 2x and leg2 = 3x

∵ leg1 = 2x and leg2 = 3x

→ Substitute them in the rule of the area above

∴ 12 = \frac{1}{2} × 2x × 3x

∵ 2x × 3x = 6x²

∴ 12 =  \frac{1}{2} × 6x²

∴ 12 = 3x²

→ Divide both sides by 3 to find x²

∴ 4 = x²

→ Take √ for both sides

∴ x = 2

→ Substitute x in the expressions of leg1 and leg2 to find them

∴ leg1 = 2(2) = 4 inches

∴ leg2 = 3(2) = 6 inches

∵ hypotenuse = \sqrt{(leg1)^{2}+(leg2)^{2}}

∴ hypotenuse = \sqrt{(4)^{2}+(6)^{2}}=\sqrt{16+36}=\sqrt{52}

∵ The simplest form of \sqrt{52} = 2\sqrt{13}

∴ The length of the hypotenuse = 2\sqrt{13} inches

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2 years ago
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dmitriy555 [2]
It's shust the opposite
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