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Andreyy89
2 years ago
13

A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(

t), in feet, at different times, t, in seconds:
f(t) = −16t2 + 32t + 384

The average rate of change of f(t) from t = 4 seconds to t = 6 seconds is :blank: feet per second.
Mathematics
2 answers:
Alex_Xolod [135]2 years ago
6 0
The answer for the blank is simply -128
neonofarm [45]2 years ago
3 0
For this case we have the following function:
 f (t) = -16t2 + 32t + 384
 By definition we have to:
 The average rate of change:
 AVR = (f (t2) - f (t1)) / (t2 - t1)
 Where,
 For t1 = 4
 f (4) = -16 * (4) ^ 2 + 32 * (4) + 384
 f (4) = 256
 For t2 = 6
 f (6) = -16 * (6) ^ 2 + 32 * (6) + 384
 f (6) = 0
 Substituting values we have:
 AVR = (0 - 256) / (6 - 4)
 AVR = -128
 Answer:
 
The average rate of change of f (t) from t = 4 seconds to t = 6 seconds is -128 feet per second.
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we know that

Angles in a linear pair are supplementary angles

So

Let

A-------> the first angle

B------> the second angle

A+B=180\\ ----> equation 1

A=13x+7\\ B=2A-1=B=26x+13

Substitute the values of A and B in the equation 1

13x+7+26x+13=180\\ 39x=180-20\\ x=\frac{160}{39}

\frac{160}{39}=4.10\ degrees

therefore

the answer is

the value of x is 4.10\ degrees


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A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto
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\begin{cases}y=2x-2\\y=2x+2\end{cases}\implies\begin{cases}-2x+y=-2\\-2x+y=2\end{cases}

For these lines, let u=-2x+y.

\begin{cases}y=2-x\\y=4-x\end{cases}\implies\begin{cases}x+y=2\\x+y=4\end{cases}

And for these, let v=x+y.

Now,

\begin{cases}u=-2x+y\\v=x+y\end{cases}\implies \begin{bmatrix}u\\v\end{bmatrix}=\underbrace{\begin{bmatrix}-2&1\\1&1\end{bmatrix}}_{\mathbf T}\begin{bmatrix}x\\y\end{bmatrix}

The vertices of S in the x-y plane are (0, 2), (2/3, 10/3), (2, 2), and (4/3, 2/3). Applying \mathbf T to each of these yields, respectively, (2, 2), (2, 4), (-2, 4), and (-2, 2), which are the vertices of a rectangle whose sides are parallel to the u-v plane.
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2 years ago
The engineers designing the All Aboard Railroad between Boca Raton and Jupiter decide to create parallel tracks through this por
olga2289 [7]

Answer:

there are no signs between the x and y and constant

it could be

2x+5y=15

2x+5y=-15

-2x+5y=15

2x-5y=15

for ax+by=c, the equation of a line paralell to that is

ax+by=d where a=a, b=b, and c and d are constants

(for this answer, I'm going to use 2x+5y=15)

given 2x+5y=15, the equation of a line paralell to that is 2x+5y=d

to find d, subsitute the point (4,-2), basically put 4 in for x and -2 for y to get the constant

2x+5y=d

2(4)+5(-2)=d

8-10=d

-2=d

the eqaution is 2x+5y=-2 (Only if the original equation is 2x+5y=-15

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5 0
2 years ago
If the legs of an isosceles right triangle have a length of 15 StartRoot 2 EndRoot ft, what is the length of the hypotenuse? 7.5
Likurg_2 [28]

Answer:

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Step-by-step explanation:

Here, we are to calculate the length of the hypotenuse of an iscosceles right triangle.

For the triangle to be isosceles, it means that the opposite and the adjacent are equal in length.

Now, to calculate the value of the hypotenuse, we make use of the Pythagoras’ theorem which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Mathematically;

let’s say the hypotenuse is length h feet

h^2 = {15√(2)}^2 + {15√(2)}^2

h^2 = (225 * 2) + (225 * 2)

h^2 = 450 + 450

h^2 = 900

h = √(900)

h = 30 feet

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2 years ago
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