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Minchanka [31]
4 years ago
7

Peter walks 30 feet away from his house and places a mirror on the ground. He backs 6 feet away from the mirror so that he can s

ee the tip of the roof. Peter's eyes are 5 feet above the ground. Peter and the house are both perpendicular to the ground. The angles between the top of the house, the mirror, and the ground and between Peter's eyes, the mirror, and the ground are congruent as shown in the image below:
Mathematics
1 answer:
nika2105 [10]4 years ago
8 0
25 is the correct answer

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Marcus and Judy are picking apples at the end of the day Marcus has a apples, Judy has five times as many apples as Marcus how m
vazorg [7]

Answer:

5<em>a</em>

Step-by-step explanation:

We know that Judy picked up 5 times as many apples as Marcus. We also know that Marcus has <em>a</em> apples. In order to find out how many Judy has, we simply multiple how many apples Marcus has by 5. When we do that, we get

5 x <em>a </em>= 5<em>a</em>

5 0
3 years ago
A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
Verdich [7]

Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

5 0
4 years ago
Please help with my last maths question for today
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Rectangle-circle
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Complete the flow proof below by filling in the reasons for 1, 2, 3, 4, and 5.
Andrei [34K]

Answer:

did u ever get the answer i need helpppp

Step-by-step explanation:

8 0
3 years ago
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a diver is swimming at -10 m. He then descends 3m and rises 6m. At what new level is the diver swimming?
iVinArrow [24]
So we start at -10. Because he descends 3 feet, We subtract 3, getting -13. He then rises 6 feet. -13 + 6 = -7. The diver is now at -7 feet. 
7 0
3 years ago
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