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Alexxandr [17]
3 years ago
13

the lines below are perpendicular. if the slope of the green lines is 2/3 what is the slope of the red line

Mathematics
1 answer:
wariber [46]3 years ago
4 0
If two lines are perpendicular then they have slopes that are negative reciporacols of eachother.

If m_1=\frac{2}{3} then m_2=-\frac{3}{2}
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Read 2 more answers
Ball kicked into the air with an initial upward velocity of 60 ft/s. It’s height h in feet after t seconds is given by the funct
Free_Kalibri [48]

Answer:

Max height: 61.25 feet

Max height reached at 15/8 seconds, or 1.875 seconds

How long to hit the ground: (15 + 7√5)/8  or about 3.83 seconds

How high after 1 second:  49 feet

Step-by-step explanation:

Since the function is a quadratic representing height, and the coefficient of the t² is negative, the vertex of the parabola will be the maximum height achieved by the ball.  

The general form for a quadratic equation is ax² + bx + c,

 here a is -16, and b is 60

 

To find the x coordinate of the vertex, use   x = -b/(2a)

  We have x = -60/[2(-16)]

                x = -60/-32

                         x = 15/8

So at 15/8 seconds, the ball reaches is maximum height

Now plug that into the equation to find the y value, which will be the height...

y = -16(15/8)² + 60(15/8) + 5  

      y = -16(225/64) + 900/8 + 5

     y = -225/4 +  450/4 + 20/4

       y =  245/4

           y = 61.25 feet

To find out how long the ball was in flight, solve the equation...

0 = -16t² + 60t + 5

    Use quadratic equation...

    x = -60/-32 ± √[60² - 4(-16)(5)]/-32

     x = 15/8 ± (√3920)/-32

      x = 15/8 ± (28√5)/-32

          x = 15/8 ± (-7√5)/8

  so x = (15 - 7√5)/8   and  (15 + 7√5)/8

     (15 - 7√5)/8 is negative, and we're talking about time, so this answer is ignored.  

(15 + 7√5)/8 seconds is when the ball hits the ground

To find out out how high the ball was after 1 second, plug 1 in for x and simplify

h(1) = -16(1²) + 60(1) + 5

 h(1) = -16 + 60 + 5

  h(1) = 49

4 0
3 years ago
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