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denis23 [38]
3 years ago
15

Find the area of the figure shown below and type your result in the empty box.

Mathematics
2 answers:
Vadim26 [7]3 years ago
7 0

Answer:

150

Step-by-step explanation:

The are is the sum of are of square and area of triangle.

A=A1+A2

A1=a*a=10*10=100

A2=a*h/2=10*10/2=50

A=100+50=150

soldier1979 [14.2K]3 years ago
5 0
10x10=100 (square) 10x10/2=50 (triangle) 100+50=150
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Which expression represents a number that is 4 times as great as the quotient of 25 and 5? A. ( 25 × 5 ) × 4 B. 4 ÷ ( 25 × 5 ) C
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Answer:

It would be D.

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A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of
Ivenika [448]

The equation for the height of the rocket at time t given

h= -16t^2+192t

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

h= -16t^2+192t

560 = -16t^2+192t

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

560 = -16(t^2 - 12t)

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

560/-16 = -16(t^2-12t)/-16

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Now we will move -35 to the righ side by adding 35 to both sides.

-35+35 = t^2-12t+35

0 = t^2 -12t+35

t^2-12t+35 = 0

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

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(t-5)(t-7) =0

So by using zero product property we will get

t-5 =0

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Also t-7 =0

t-7+7 = 0+7

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So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

h= -16t^2+192t

0= -16t^2 + 192 t

0 = -16(t^2-12t)

-16(t^2-12t) = 0

We will move -16 to the other side by dividing it to both sides.

-16(t^2-12t)/-16 = 0/-16

t^2-12t = 0

We will take out the common factor t from the left side. By taking out t we will get,

t(t-12) = 0

We will use zero product property now. By using that we will get,

t = 0

ans also t-12 = 0

t-12+12 = 0+12

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When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

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