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EastWind [94]
2 years ago
7

1. The angle of depression from the top of the

Mathematics
1 answer:
STatiana [176]2 years ago
3 0

Answer:

41.7feet

Step-by-step explanation:

From the question we are given the following

angle of depression = 50°

Distance of the pole from the base of the feet = 35feet (Adjacent)

Required

height of the school (opposite)

Using the SOH CAH TOA identity

Tan theta = opp/adj

Tan 50 = H/35

H = 35tan 50

H = 35(1.1918)

H = 41.7feet

Hence the height of the school is 41.7feet

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The base of a parallelogram is 18 inches longer than three times the height. The area of the parallelogram is 216 square inches.
nordsb [41]

Answer:

12 inches

Step-by-step explanation:

Let b represent the base

h represents the height

area of the parallelogram = base * height = 216 square inches

From the question'

b = 18 + 3h

Slot in the value of b

216 = (18 + 3h) * h

expand

216 = 18h + 3h^2

subtract 216 from both sides

0 = 18h + 3h^2 - 216

rearrange

3h^2 + 18h - 216 = 0

divide through by 3

h^2 + 6h - 72 = 0

Now, lets solve!

h^2 + 6h - 12h - 72 = 0

h( h + 6 ) - 12(h + 6) = 0

(h - 12) (h + 6) = 0

h - 12 = 0

h = 12

and

h + 6 = 0

h = - 6

Taking the positive value of h

Hence, the height is 12 inches

Lets check

when h = 12 inches

Area of the parallelogram = 18* 12 = 216 square inches     .... correct

when h = -6inches

A = 18 * -6 ≠  216 square inches

So height is 12 inches

5 0
3 years ago
How to solve derivative of (sin3x)/x using first principle ​
Leona [35]

\dfrac{d}{dx}(\dfrac{\sin(3x)}{x})

First we must apply the Quotient rule that states,

(\dfrac{f}{g})'=\dfrac{f'g-g'f}{g^2}

This means that our derivative becomes,

\dfrac{\dfrac{d}{dx}(\sin(3x))x-\dfrac{d}{dx}(x)\sin(3x)}{x^2}

Now we need to calculate \dfrac{d}{dx}(\sin(3x)) and \dfrac{d}{dx}(x)

\dfrac{d}{dx}(\sin(3x))=\cos(3x)\cdot3

\dfrac{d}{dx}(x)=1

From here the new equation looks like,

\dfrac{3x\cos(3x)-\sin(3x)}{x^2}

And that is the final result.

Hope this helps.

r3t40

7 0
3 years ago
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solmaris [256]
The Max amount of times he can go is 5 times
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A 3-yard-long ribbon was used to trim for dresses. Each dress used the same amount of ribbon. How much ribbon was used for each
Tcecarenko [31]

Answer:

For x number of dresses, the answer is 3/x.

Step-by-step explanation:

Divide the length of the ribbon by the number of dresses.

If there were x dresses, the answer is 3/x.

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How many groups of Two-thirds are in Three-fifths?
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Answer:

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Did the test

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