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Reptile [31]
2 years ago
13

A model rocket is launched with an initial upward velocity of 138 ft/s. The rocket's height h (in feet) after 1 seconds is given

by the following.
h = 138t-16t^2
Find all values of 1 for which the rocket's height is 78 feet.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
Mathematics
1 answer:
insens350 [35]2 years ago
8 0

Answer:

t = 8.017 or 0.608 (nearest hundredth)

Step-by-step explanation:

h = 138t-16t^2

when h = 78:

\implies 138t-16t^2=78

\implies 16t^2-138t+78=0

Using quadratic formula:

t=\dfrac{138 \pm\sqrt{(-138)^2-(4 \cdot 16 \cdot 78)} }{2 \cdot 16}

\implies t=\dfrac{138 \pm\sqrt{14052} }{32}

\implies t=8.017, t = 0.608

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Trava [24]

9514 1404 393

Answer:

  x = 10

Step-by-step explanation:

The marked sides are proportional, so we have ...

  x/8 = (x +5)/12

  3x = 2(x +5) . . . . . multiply by 24 to clear fractions

  3x = 2x + 10 . . . . . eliminate parentheses

  x = 10 . . . . . . . . . . subtract 2x

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<em>Alternate solution</em>

We observe that the difference between the upper segment lengths is ...

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and the difference between the lower segment lengths is ...

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3 years ago
24 : 18<br> In its simplest form
beks73 [17]

Answer:

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Step-by-step explanation:

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3 years ago
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Solve the equation 4s2+32s=60 by completing the square
lord [1]

Answer:

s = -4 ± √31

Step-by-step explanation:

Given 4s^{2} + 32s = 60

divide by 4 throughout

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3 years ago
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Andrei [34K]

Answer:

radius = \sqrt{13} or radius = 3.61

Step-by-step explanation:

Given

Points:

A(-3,2) and B(-2,3)

Required

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First, we have to determine the center of the circle;

Since the circle has its center on the x axis; the coordinates of the center is;

Center = (x,0)

Next is to determine the value of x through the formula of radius;

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Considering the given points

A(x_1,y_1) = A(-3,2)

B(x_2,y_2) = B(-2,3)

Center(x,y) =Center (x,0)

Substitute values for x,y,x_1,y_1,x_2,y_2 in the above formula

We have:

\sqrt{(-3 - x)^2 + (2 - 0)^2} = \sqrt{(-2 - x)^2 + (3 - 0)^2}

Evaluate the brackets

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

\sqrt{(-(3 + x))^2 + 4} = \sqrt{(-(2 + x))^2 + 9}

Eva;uate all squares

\sqrt{(-(3 + x))(-(3 + x)) + 4} = \sqrt{(-(2 + x))(-(2 + x)) + 9}

\sqrt{(3 + x)(3 + x) + 4} = \sqrt{(2 + x)(2 + x) + 9}

Take square of both sides

(3 + x)(3 + x) + 4 = (2 + x)(2 + x) + 9

Evaluate the brackets

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Substitute 0 for x

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Considering that we used x1 and y1;

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Substitute -3 for x1 and 2 for y1

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