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Gennadij [26K]
2 years ago
13

Please help me!! I need to get this question

Mathematics
1 answer:
Sergio [31]2 years ago
7 0

Answer:

Part A;

- Increases

- Increases

Part B:

- Constant but not 0

- 0

Step-by-step explanation:

Part A:

>> Looking at the speed - time graph of Airplane A, we can see that speed increases from rest at 0 seconds time to time.

>> In graph of airplane B, we can see that distance gradually increases from time at 0 to time at t1. This means that the speed is increasing.

Part B:

>> From time t1 to time t2, we see in airplane A graph that the motion is indicated a horizontal line which means there is constant motion and in constant motion, speed is constant but not zero.

>> In graph of airplane B, we see that the distance between time t1 to time t2 didn't increase. This means that there was no movement. As a result, speed will be 0

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Please answer this the best you possible can :)
Veronika [31]

Answer:

the answer is x=15

Step-by-step explanation:

75/5=15

120/8=15

180/12=15

Hope this helps some.

7 0
3 years ago
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Find the measure of < AED for m < BEC = 96
Strike441 [17]
I wish you would of put a picture but with what you gave me. if the angle is a straight line then you would just set it up like 96+x=180 bc a straight line ='s 180 and den subtract 96 from 180 which is 84 so x would = 84 if the angle is a supp. angle but til there's not a picture this might be wrong.
But i hope this helps have a nice nite!
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3 years ago
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What is the effect on the graph of f(x) = x2 when it is transformed to h(x) = 2 x2 + 15?
kirill [66]

Answer:

B

Step-by-step explanation:

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3 years ago
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Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
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3 years ago
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