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2 years ago

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3. MATCH is a regular pentagon with coordinates M(-1,1) and A(3,0). What is the perimeter of MATCH?

1 answer:

Makovka662 [10]2 years ago

6 0

Answer:

<h3>2.squareroot 85</h3>

Step-by-step explanation:

<h2>#carry on learning</h2><h2>#mark branlits</h2>

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Answer:

There is one Zero.

Step-by-step explanation:

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6⋅7-3^2⋅9+4^3 what is the value?

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$\textsf{Hey there!}$

$\mathsf{Equation: 6 \times 7- 3^2 \times9+4^3}$

$\mathsf{3^2 = 3\times3=9}$

$\mathsf{4^3= 4\times4\times4=16\times4 = 64}$

$\mathsf{6\times7 = 42}$

$\mathsf{9\times9= 81}$

$\mathsf{New\ equation: 42 - 81 + 64}$

$\mathsf{42 -81= -39}$

$\textsf{New equation: -39 + 64}$

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NEED HELP NOW 100 POINTS WILL MARK BRAINLIEST PLZ ANSWER ALL 5 PHOTOS BELOW!!!!!!

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The answers are as follows:
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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.

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

$-35 = t^2 -12t$

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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So by using zero product property we will get

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Also $t-7 =0$

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

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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.

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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 $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

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.

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