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marishachu [46]
3 years ago
12

The height in feet of a water rocket launched from the ground is given by the function h(t) = -16t^2 + 96t where t represents th

e number of seconds after launch. What is the maximum heigh attained by the rocket?
Mathematics
2 answers:
Mazyrski [523]3 years ago
7 0
<span>The biggest height is 144 ft at t = 3 seconds.</span>
Margarita [4]3 years ago
7 0
Let's imagine this function in real life. We the rocket is launched, it will follow a path similar to a negative parabola (a 'frown', as the coefficient of x^2 is negative).

So, we can work out the answer by finding the vertex of this of this 'graph', as this represents the highest point that the rocket will reach during its journey. To find t, we use the expression -b/2a (these letters are from the format of the quadratic equation ax^2+bx+c, so in this case a= -16 and b= 96):
-b/2a
-(96)/2(-16)
-96/-32
t=3

Now, we can plug 3 back into the function:
h(t)= -16t^2+96t
h(3)= -16(3^2)+96(3)
h(3)= -16(9)+96(3)
h(3)= -144+288
h(3)= 144

Therefore, the maximum height attained by the rocket is 144ft, at t = 3 seconds.
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If KX - 8Y + 4 =0 is parallel to the line who’s equation is 3X -4Y+8 = 0. What is the value of K
Alex_Xolod [135]

So - first you gotta put it in the y = mx + b form.


4Y = 3X +8

=> y = 4/3X + 1/2

A parallel line has the same slope. So if 3X - 4Y + 8=0 has a slope of 4/3 then the value of K must also be 4/3

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=-3%5Csqrt%7B45y%C2%B3%7D" id="TexFormula1" title="-3\sqrt{45y&sup3;}" alt="-3\sqrt{45y&sup3;}"
GarryVolchara [31]

Answer:

-9A · √(5yA)

Step-by-step explanation:

The coefficient -3 stays the same.

45 factors into 5·9, which is helpful because 9 is a perfect square.

Thus, √45 = 3√5.

y cannot be factored.  It stays under the radical.

A³ can be factored into A² (a perfect square) and A.

Thus,

-3√(45yA³) = -3 · 3√5 · √y · A · √A, or

                   = (-3)(3)(A) · √(5yA), or

                    = -9A · √(5yA)

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1 : 40 = 6 : 240

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