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Amanda [17]
3 years ago
11

Does anyone know this one?

Mathematics
2 answers:
34kurt3 years ago
8 0

1. Multiply both sides by 2 ----------> 2R = l + 3w

2. Subtract l from both sides ----> 2R - l = 3w

3. Divides both sides by 3 ----> (2R - l) / 3 = w

seropon [69]3 years ago
6 0

Answer:

Multiply both sides by 2

Subtract l from both sides

Divide each side by 3

(2R -l)/3 =w

Step-by-step explanation:

R = (l+3w)/2

Multiply both sides by 2

2R = 2* (l+3w)

2R = l+3w

Subtract l from both sides

2R - l = l-l +3w

2R -l = 3w

Divide each side by 3

(2R-l) /3 = 3w/3

(2R -l)/3 =w

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Which of the following statements is true about an image after a dilation?
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Answer:

Not sure what the answer choices are, but choose the choice that says the new image is either stretched or shrunk. In a dilation, the shape/corresponding sides of the pre-image are preserved in the new image, but the size of the new image is altered.

Step-by-step explanation:

hope this helps!

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3 years ago
If the area of square is 2401 sqaure metre,find the perimeter of the sqaure​
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P=196                        

P=4A=4·2401=196                                  

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3 years ago
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a girl scout has (5x-12) boxes of cookies and sells (3x+18) of them. Write a simplified expression to represent the number of bo
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5x-12 -(3x+18)

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3 years ago
A pole that is 2.8 m tall casts a shadow that is 1.21 m long. At the same time, a nearby tower casts a shadow that is 48.75 m lo
DanielleElmas [232]

Answer:

112.81 m.

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8 0
3 years ago
Simplify the expression. tan(sin^−1 x)
Blizzard [7]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2799412

_______________


Let  \mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}

(that is the range of the inverse sine function).


So,

\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}


Square both sides:

\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}


Since \mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},} then \mathsf{cos\,\theta} is positive. So take the positive square root and you get

\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}


Then,

\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}


I hope this helps. =)


Tags:  <em>inverse trigonometric function sin tan arcsin trigonometry</em>

3 0
3 years ago
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