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stiv31 [10]
3 years ago
8

NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=-4.9t2+64t+

136.
The rocket will reach its peak height of 345 meters above sea level at _____ second
Mathematics
1 answer:
ryzh [129]3 years ago
7 0

Using the vertex of the quadratic equation, it is found that:

The rocket will reach its peak height of 345 meters above sea level at 6.53 seconds.

<h3>What is the vertex of a quadratic equation?</h3>

A quadratic equation is modeled by:

y = ax^2 + bx + c

The vertex is given by:

(x_v, y_v)

In which:

x_v = -\frac{b}{2a}

y_v = -\frac{b^2 - 4ac}{4a}

Considering the coefficient a, we have that:

  • If a < 0, the vertex is a maximum point.
  • If a > 0, the vertex is a minimum point.

In this problem, the equation is given by:

h(t) = -4.9t² + 64t + 136.

The coefficients are a = -4.9 < 0, b = 64, c = 136, hence the instant of the maximum height is given by, in seconds:

t_v = -\frac{64}{2(-4.9)} = 6.53

More can be learned about the vertex of a quadratic equation at brainly.com/question/24737967

#SPJ1

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qaws [65]
There are 13 angles.
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1.

m\angle2+m\angle3+m\angle4=180\qquad\qquad\text{linear angles}\\\\m\angle3+m\angle3+m\angle4=180\qquad\qquad m\angle2=m\angle3\\\\2m\angle3+m\angle4=180\\\\2m\angle3+112=180\\\\2m\angle3=180-112\\\\2m\angle3=68\quad|:2\\\\\boxed{m\angle3=34=m\angle2}

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2.

m\angle3+m\angle12+m\angle11=180\qquad\qquad\text{triangle}\\\\34+105+m\angle11=180\\\\139+m\angle11=180\\\\m\angle11=180-139\\\\\boxed{m\angle11=41}

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3.

m\angle12+m\angle13=180\qquad\qquad\text{linear angles}\\\\105+m\angle13=180\\\\m\angle13=180-105\\\\\boxed{m\angle 13=75}

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4.

m\angle1+m\angle2+m\angle13=180\qquad\qquad\text{triangle}\\\\m\angle1+34+75=180\\\\m\angle1+109=180\\\\m\angle1=180-109\\\\\boxed{m\angle1=71}

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m\angle9=m\angle11\qquad\qquad\text{vertical angles}\\\\\boxed{m\angle9=41}

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m\angle10+m\angle11=180\qquad\qquad\text{linear angles}\\\\m\angle10+41=180\\\\m\angle10=180-41\\\\\boxed{m\angle10=139}

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