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elena-s [515]
3 years ago
13

Jamar is playing a video game. The object of the game is to roll a marble into a target. In the figure, the

Mathematics
2 answers:
OleMash [197]3 years ago
3 0

Answer:

Jamar should hit the wall opposite to the side of the barrier(if there is a space), which will cause the ball to bounce off the wall at an angle and into Target.

Brrunno [24]3 years ago
3 0

Answer: (0,2)

Step-by-step explanation:

You have to get from M to T without touching the barrier. What you do is you reflect M across the edge of the screen (y axis) and you should get (0,-1). After you do that connect T to M’ and the y intersect is your answer, (0,2)

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What is<br> 4 hr 51 min 19 sec<br> +2 hr 53 min 23 sec
Flauer [41]

224 hours and 43 seconds.

7 0
3 years ago
Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>​
Alla [95]

\large \mathbb{PROBLEM:}

\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}

\large \mathbb{SOLUTION:}

\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}

\boxed{ \tt   \red{C}arry  \: \red{ O}n \:  \red{L}earning}  \:  \underline{\tt{5/13/22}}

4 0
2 years ago
(a) The proportion of accidents that occur up to mile 1.2 of the path is the area under the density curve between 0 miles and 1.
inn [45]

Answer:

the area would be 0.4 or 40%

Step-by-step explanation:

The computation of the area is given below:

But before that the level of the density is

The density level is

=  1 ÷ (3 - 0)

= 1 ÷ 3

Area from 0 miles to 1.2 miles

= (1 ÷ 3) × (1.2 - 0)

= 0.4 or 40 %

Hence, the area would be 0.4 or 40%

the same is considered and relevant too

4 0
3 years ago
If p is a true statement and q is false, what is the truth value of p ∨ q?
Dafna1 [17]
p\lor q is true if either of p or q is true. We know p is true, which enough to say that p\lor q is also true.
3 0
3 years ago
Read 2 more answers
CAN SOMEBODY HELP ME WITH THIS QUESTION
pickupchik [31]

Answer:

no sorry

Step-by-step explanation:

I physically can't help but in ban

4 0
3 years ago
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