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3 years ago

Select all equations that have 5 as a solution.

Mathematics

1 answer:

Lina20 [59]3 years ago

4 0

Answer:

D,F

Step-by-step explanation:

D . x cubed is 5 times 5 which is 25 times 5 which is 125

F x squared is 5 times 5 which is 25

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The automobile drove 50 mph and had gone 200 miles when the airplane set out in pursuit. It took 4 hours to overtake the car. Wh

Alexxx [7]

100 mph per hour

Step-by-step explanation:

solution

speed of auto mobile=50mph

distance already traveled when the airplane took out in pursuit=200miles

time taken to overtake the car=4 hours

Now,

let the speed of airplane be X

X=50×4+200

or,X=200+200

or,X=400

or, X=400÷4

Therefore, X=100

3 0

2 years ago

Please Help!!

goldenfox [79]

Question 5 is B
Question 9 is C

5 0

3 years ago

Read 2 more answers

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.

$t^2-12t+35 =0$

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

So by using zero product property we will get

$t-5 =0$

$t-5+5 = 0+5$

$t=5$

Also $t-7 =0$

$t-7+7 = 0+7$

$t=7$

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 $t-12 = 0$

$t-12+12 = 0+12$

$t = 12$

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.

So we have got the required answers.

6 0

3 years ago

I REALLY NEED HELP ON THIS QUICK ASAP

oksian1 [2.3K]

Is there like any image to see how big are the posts?

8 0

3 years ago

What are the steps to get 1/7

madreJ [45]

1/7 is your answer. It can't be reduced any further. If there is supposed to be a problem, next time, add it.

8 0

3 years ago

Select all equations that have 5 as a solution.​

Select all equations that have 5 as a solution.