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yarga [219]
3 years ago
11

Can some body help me my mom won’t let me sleep until I find it

Mathematics
1 answer:
balu736 [363]3 years ago
5 0

OT = 4

PT = 6.93

<u>Explanation :</u>

<u />

OS = OT = R = 4

Hence ∠OTS = ∠OST, (Angles opposite to equal sides are equal)

∠TOS + ∠OTS + ∠OST = 180°

60° + ∠OTS + ∠OST = 180°

2∠OTS = 180° - 60° (∠OTS = ∠OST)

Hence, ∠TOS = ∠OTS = ∠OST = 60°

Hence, ΔOST is an equilateral triangle,

Therefore, OS = OT = ST = 4

Let us draw a perpendicular from OD on PT.

ΔODT is a right angled triangle

∠DTO = 90° - ∠OTS = 30° (Angle of a rectangle is 90°)

in ΔDTO,

cos 30° = DT/OT

\sqrt{3}/2 = DT /4

DT =2\sqrt{3} =  3.46

PT = 2*DT = 6.93

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The three trapezoids have the same base lengths, but different angle measures. Which trapezoid has the greatest area? Explain.
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Steps I took:

I drew out a circle with a radius of 1 and drew a trapezoid inscribed in the top portion of it. I outlined the rectangle within the trapezoid and the two right triangles within it. This allowed me to come to the conclusion, using the pythagorean theorem, that the height of the trapezoid is <span><span>h=<span><span>1−<span><span>x2</span>4</span></span><span>−−−−−</span>√</span></span><span>h=<span>1−<span><span>x2</span>4</span></span></span></span>

The formula for the area of a trapezoid is <span><span>A=<span><span>a+b</span>2</span>(h)</span><span>A=<span><span>a+b</span>2</span>(h)</span></span> I am basically solving for the <span>aa</span> here and so far I have:

<span><span>A=<span><span>x+2</span>2</span>(<span><span>1−<span><span>x2</span>4</span></span><span>−−−−−</span>√</span>)</span><span>A=<span><span>x+2</span>2</span>(<span>1−<span><span>x2</span>4</span></span>)</span></span>

I simplified this in order to be able to easily take the derivative of it as such:

<span><span>A=<span><span>x+2</span>2</span>(<span><span><span>4−<span>x2</span></span>4</span><span>−−−−−−</span>√</span>)</span><span>A=<span><span>x+2</span>2</span>(<span><span>4−<span>x2</span></span>4</span>)</span></span>

<span><span>A=<span><span>x+2</span>2</span>(<span><span><span>4−<span>x2</span></span><span>−−−−−</span>√</span>2</span>)</span><span>A=<span><span>x+2</span>2</span>(<span><span>4−<span>x2</span></span>2</span>)</span></span>

<span><span>A=(<span>14</span>)(x+2)(<span><span>4−<span>x2</span></span><span>−−−−−</span>√</span>)</span><span>A=(<span>14</span>)(x+2)(<span>4−<span>x2</span></span>)</span></span>

Then I took the derivative:

<span><span><span>A′</span>=<span>14</span>[(1)(<span><span>4−<span>x2</span></span><span>−−−−−</span>√</span>)+(x+2)((<span>12</span>)(4−<span>x2</span><span>)<span>−1/2</span></span>(−2x)]</span><span><span>A′</span>=<span>14</span>[(1)(<span>4−<span>x2</span></span>)+(x+2)((<span>12</span>)(4−<span>x2</span><span>)<span>−1/2</span></span>(−2x)]</span></span>

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6 0
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Read 2 more answers
- Varun drove to work this morning. It took
babymother [125]

Answer:

See attached graph

Step-by-step explanation:

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Then we represent his driving at constant speed (although it has not been specified in the problem) as a line with positive slope equal to his speed, that goes on for the first five minutes. After that, he reaches the traffic light that keeps him at the same distance from home for three minutes (notice the line representing the distance covered is flat from minute 5 to minute 8, while he is not moving).

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The line ends at minute 15 which is the time it took him to get to work.

6 0
3 years ago
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