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Eva8 [605]
3 years ago
10

A skate board ramp is built on an incline so that it rests on a porch that is 1.3m high. The bottom of the ramp is placed 2.5m f

rom the porch. What is the angle that the ramp forms with the ground?

Mathematics
1 answer:
mr_godi [17]3 years ago
8 0

Answer:

θ = 27.47°

Step-by-step explanation:

This problem can be represented as a right-angled triangle, as shown in the diagram below.

To find θ, we apply the tangent part of the trigonometric ratios SOHCAHTOA:

tanθ = \frac{opposite}{adjacent}

In the diagram, opposite = 1.3 m and adjacent = 2.5 m

Hence:

tanθ = \frac{1.3}{2.5}

tanθ = 0.52

θ = tan^{-1} 0.52

θ = 27.47°

The ramp forms 27.47° with the ground.

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Answer:

y=3

Step-by-step explanation:

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30 points for the correct answer
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Answer:

its either 2/9 or 2/3

Step-by-step explanation:

I solved the top part and got 2/3 and i solved the bottom part and got 1/3 and you can multiply 2/3 and 1/3 and got 2/9

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2 years ago
Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first
BaLLatris [955]

Answer:

\cos(x+y) goes with -\frac{\sqrt{6}+\sqrt{2}}{4}

\sin(x+y) goes with \frac{\sqrt{6}-\sqrt{2}}{4}

\tan(x+y) goes with \sqrt{3}-2

Step-by-step explanation:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

We are given:

\sin(x)=\frac{\sqrt{2}}{2} which if we look at the unit circle we should see

\cos(x)=\frac{\sqrt{2}}{2}.

We are also given:

\cos(y)=\frac{-1}{2} which if we look the unit circle we should see

\sin(y)=\frac{\sqrt{3}}{2}.

Apply both of these given to:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}

\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}

\frac{-\sqrt{2}-\sqrt{6}}{4}

-\frac{\sqrt{6}+\sqrt{2}}{4}

Apply both of the givens to:

\sin(x+y)

\sin(x)\cos(y)+\sin(y)\cos(x) by addition identity for sine.

\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}

\frac{-\sqrt{2}+\sqrt{6}}{4}

\frac{\sqrt{6}-\sqrt{2}}{4}

Now I'm going to apply what 2 things we got previously to:

\tan(x+y)

\frac{\sin(x+y)}{\cos(x+y)} by quotient identity for tangent

\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

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-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}

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There is a perfect square in 12, 4.

-\frac{8-2\sqrt{4}\sqrt{3}}{4}

-\frac{8-2(2)\sqrt{3}}{4}

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Divide top and bottom by 4 to reduce fraction:

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Distribute:

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Answer:

10 lol..................

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anzhelika [568]

Answer:

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Step-by-step explanation:

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This is approximately equal to 3 when rounded up to the nearest whole number.

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