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ki77a [65]
4 years ago
12

Use the following work shown below to answer the following question.​

Mathematics
1 answer:
zysi [14]4 years ago
3 0

Answer:

\frac{3 + \sqrt{3x} }{3 - x}

Step-by-step explanation:

You can use the work they gave you and continue from there:

\frac{\sqrt{3}(\sqrt{3}) + \sqrt{3}(\sqrt{x})  }{(\sqrt{3}) ^{2} - (\sqrt{x} )^{2} }

First, simplify the numerator:

\frac{3 + \sqrt{3x} }{(\sqrt{3}) ^{2} - (\sqrt{x}) ^{2} }

Finally, simplify the denominator:

\frac{3 + \sqrt{3x} }{3 - x}

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Llana [10]

Answer:

a. h = 60t − 4.9t²

b. 12.2 seconds

c. 183.7 meters

Step-by-step explanation:

a. Given:

y₀ = 0 m

v₀ = 60 m/s

a = -9.8 m/s²

y = y₀ + v₀ t + ½ at²

h = 0 m + (60 m/s) t + ½ (-9.8 m/s²) t²

h = 60t − 4.9t²

b. When the ball lands, h = 0.

0 = 60t − 4.9t²

0 = t (60 − 4.9t)

t = 0 or 12.2

The ball lands after 12.2 seconds.

c. The maximum height is at the vertex of the parabola.

t = -b / (2a)

t = -60 / (2 × -4.9)

t = 6.1 seconds

Alternatively, the maximum height is reached at half the time it takes to land.

t = 12.2 / 2

t = 6.1 seconds

After 6.1 seconds, the height reached is:

h = 60 (6.1) − 4.9 (6.1)²

h = 183.7 meters

7 0
3 years ago
Please help me.
sammy [17]
What grade is this problem for?
6 0
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Write an inequality and solve each problem.For exercises 11 and 12, interpret the solution.
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Help please :( I’m stuck on this..
Diano4ka-milaya [45]

Answer:

4x10+1

Step-by-step explanation:

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