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3 years ago

BRAINLIESTTTT ASAP! PLEASE ANSWER

Mathematics

2 answers:

iren [92.7K]3 years ago

6 0

H(t) = −16t^2 + 75t + 25
g(t) = 5 + 5.2t

A)
At 2, h(t) = 111, g(t) = 15.4
At 3, h(t) = 106, g(t) = 20.6
At 4, h(t) = 69, g(t) = 25.8
At 5, h(t) = 0, g(t) = 31
The heights of both functions would have been the closest value to each other after 4 seconds, but before 5 seconds. This is when g(x) is near 30 (26-31), and the only interval that h(t) could be near 30 is between 4 and 5 seconds (as it is decreasing from 69-0).

B) The solution to the two functions is between 4 and 5 seconds, as that is when their height is the same for both g(t) and h(t). Actually the height is at 4.63 seconds, their heights are both
What this actually means is that this time and height is when the balls could collide; or they would have hit each other, given the same 3-dimensional (z-axis) coordinate in reality.

meriva3 years ago

6 0

Answer and Explanation :

We have given that, The function shows the height H(t) $H(t)=-16t^2+75t+25$ , in feet, of a projectile after t seconds.

A second object moves in the air along a path represented by $g(t) = 5 + 5.2t$ , where g(t) is the height, in feet, of the object from the ground at time t seconds.

We have to find :

Part A - Create a table using integers 2 through 5 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located?

To create a table substitute the value of t from 2 to 5 in the equation H(t) and g(t) we get,

t $H(t)=-16t^2+75t+25$ $g(t) = 5 + 5.2t$

2 $-16(2)^2+75(2)+25=111$ $5 + 5.2(2)=15.4$

3 $-16(3)^2+75(3)+25=106$ $5 + 5.2(3)=20.6$

4 $-16(4)^2+75(4)+25=69$ $5 + 5.2(4)=25.8$

5 $-16(5)^2+75(5)+25=0$ $5 + 5.2(5)=31$

Now, To find the the interval in which H(t)=g(t) we have to plot the equations and the points.

We have seen that the intersection point of both the equation is (4.632,29.088) and it lies between 4 to 5.

So, The solution to H(t) = g(t) located between t=4 and t=5.

Refer the attached figure below.

Part B - Explain what the solution from Part A means in the context of the problem.

Part A signifies that the projectile motion of the equations are same in the interval (4,5).

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Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$