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arlik [135]
3 years ago
15

an eagle leaves her nest on the side of a cliff. she soars upward 60 ft and then dives 80 ft. what is her change in elevation af

ter leaving the nest
Mathematics
1 answer:
labwork [276]3 years ago
5 0

Her change is -20ft.

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Use the following long division.
Kay [80]

Answer:

A. X^2

Step-by-step explanation:

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3 years ago
Which value of b would make x²+bx-30 factorable?<br> A. -31<br> B. -17<br> C. 11<br> D. 13
dimulka [17.4K]

Answer:

D. 13

Step-by-step explanation:

well since it's a quadratic equation, it can be written in the form of ax²+bx+c=0. So we have a=1, b=b and c= -30. Also the roots are given by the equation-

x= -b±\sqrt{b^{2}-4ac} /2a. So all I did was put the number from the choices which would satisfy the sqrt part and that turned out to be 13 since \sqrt{13^{2}-4*1*(-30) = \sqrt{289} = 17. So b=13.

5 0
3 years ago
Read 2 more answers
Plz help me with this
viva [34]

Answer:

The height on the plans is 6.5 inches

Step-by-step explanation:

1 inch of plan = 5 inch of dollhouse

we know the dollhouse is 32.5 inches, so we need to determine the unknown

x inch of plan = 32.5 inch of dollhouse

x = 32.5/5 = 6.5 inches

8 0
3 years ago
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
Someone please tell me how to solve this
rewona [7]
The answer will be 21u^6v^8
6 0
3 years ago
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