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OverLord2011 [107]
3 years ago
11

Pam and Jim were inspired by the most recent Space Xrocket launch (pictured below) and decided to build a rocket of their own. A

fter
several trials, they built a rocket that, when launched, followed a path that looks like a parabola,
It was important to them to know exactly where it landed each time so they could retrieve the rocket for parts, On their most recent
launch, Pam used a computer to record the path of the rocket. The computer reported the path the rocket took followed this equations
y = -4x^2 + 48xwhere y = height of the rocket in yards and X - horizontal distance in yards,
How far away from the launch point did the rocket land?
Find the x-intercepts to help answer the question, Explain how you got your answer below. Be sure to show all your calculations
supporting your answer on your paper,
Mathematics
1 answer:
Anastaziya [24]3 years ago
3 0
I can’t see that please try again
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C<br> IT<br> W<br> D<br> B 6<br> 20<br> HELP PLEASE
nikdorinn [45]

answer:

b6

step bye step?

soalnya itu dibagi w = b6 maaf kalo salah wkwk

7 0
3 years ago
Solving Equations with the Variable on Each Side · Practice
JulijaS [17]

Hello friend , I hope it's helps you

enjoy your day

4 0
3 years ago
Read 2 more answers
If anyone could help, it would mean the world. thank you
kondaur [170]
I forgot that but good luck
7 0
3 years ago
A total of 611 tickets were sold for the school play. They were either adult tickets or student tickets. There were 61 more stud
Sergio039 [100]

Answer:

Adult tickets=275

Step-by-step explanation:

form 2 equations from the given info.

x+y=611

y=x+61

x being the adults and y being the students tickets

then substitute the y equation into the first equation.

x+x+61=611

2x+61=611

2x=550

x=275

and you can check by putting 275 back into the original equations :)

7 0
3 years ago
The distance of an object thrown upward is modeled by the equation h=-2x^2+3t+4, where h is the height of the object in feet at
kipiarov [429]

So for this, we will have to find the zeros (x-intercepts) of the equation. In this case, we will be using the quadratic formula, which is x=\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a} (a = x^2 coefficient, b = x coefficient, c = constant). Using our info, our equation is such: x=\frac{-3+\sqrt{3^2-4*(-16)*4}}{2*(-16)},\frac{-3-\sqrt{3^2-4*(-16)*4}}{2*(-16)}

Firstly, solve the multiplications and the exponents: x=\frac{-3+\sqrt{9+256}}{-32},\frac{-3-\sqrt{9+256}}{-32}

Next, do the addition: x=\frac{-3+\sqrt{265}}{-32},\frac{-3-\sqrt{265}}{-32}

Next, plug in the equations into the calculator and your answer will be: x=-0.41,0.60

Since we can't have negative time in this situation, it can't be -0.41 seconds. Which means that the object is visible for about 0.60 seconds, or C.

8 0
4 years ago
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