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Vedmedyk [2.9K]
3 years ago
6

A photographer shines a camera light at a particular painting forming an angle of 40° with the camera platform. If the light is

58 feet from the wall where the painting hangs, how high above the platform is the painting?
Mathematics
1 answer:
krek1111 [17]3 years ago
4 0

Answer:

Let y be the height above the camera platform of the painting.

tan (40degrees) = y/58

y = 58[tan(40)] = 58(0.8391) = 48.67 feet.

y = 48.67 feet

Step-by-step explanation:

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LekaFEV [45]

Answer:

y = -3x + 4

Step-by-step explanation:

sorry im late

but all you have to do is isolate the y

so to do that we have to get rid of the 3x

3x + y = 4

in order to do that we have to subtract 3x from both sides of the equal sign since it is positive and being added to the y

3x + y = 4

-3x          -3x

y = -3x + 4

and now y is isolated

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Step-by-step explanation:

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nick has to build a brick wall. Each row of the wall requires 6/2 bricks. There are 10 rows in the wall. How many bricks will ni
alex41 [277]

Answer:

D. 10 x 6/2

Step-by-step explanation:

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Help me plz asap if you can
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Analysis of an accident scene indicates a car was traveling at a velocity of 69.5 mph (31.1 m/s) along the positive x-axis at th
STatiana [176]

We can find the acceleration via

{v_f}^2-{v_i}^2=2a\Delta x

We have

\left(5.20\dfrac{\rm m}{\rm s}\right)^2-\left(31.1\dfrac{\rm m}{\rm s}\right)^2=2a(115\,\mathrm m)

\implies\boxed{a=-4.09\dfrac{\rm m}{\mathrm s^2}}

Then by definition of average acceleration,

a_{\rm ave}=\dfrac{v_f-v_i}t

so that

-4.09\dfrac{\rm m}{\mathrm s^2}=\dfrac{5.20\frac{\rm m}{\rm s}-31.1\frac{\rm m}{\rm s}}t

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We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

v_{\rm ave}=\dfrac{x_f-x_i}t=\dfrac{v_f+v_i}2

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\dfrac{115\,\rm m}t=\dfrac{5.20\frac{\rm m}{\rm s}+31.1\frac{\rm m}{\rm s}}2

\implies\boxed{t=6.33\,\mathrm s}

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