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kipiarov [429]
3 years ago
11

Provide a visual of 12 to 36. Please circle the groups in order to indicate the ratio 12/36.

Mathematics
2 answers:
alexira [117]3 years ago
3 0

Answer: take a picture of the work please

Step-by-step explanation:

Ivenika [448]3 years ago
3 0
So what you do is take the 12 and add it to the 36 = ???
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A box is shaped like an octagonal prism. Here is what the base of the prism looks like. If the height of the box is 7 inches, th
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Answer:

The volume of the box is 33.796 cubic inches

Step-by-step explanation:

Given

Height = 7in

See attachment for the base of the prism

Required

The volume of the box

From the attachment, we have:

a = 1in -- side length of the octagon

The area is then calculated as:

Area = 2 * (1 + \sqrt 2)a^2

So, we have:

Area = 2 * (1 + \sqrt 2)* 1^2

Area = 2 * (1 + \sqrt 2)* 1

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Area = 2 * (2.414)

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Volume = Height * Base\ Area

Volume = 7 * 4.828

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5 0
3 years ago
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Prove that :-<br><br> Cos3A-cosA/sin3A-sinA+cos2A-cos4A/sin4A-sin2A=sinA/cos2Acos3A
ad-work [718]
\dfrac{\cos^3A-\cos A}{\sin^3A-\sin A}+\dfrac{\cos^2A-\cos^4A}{\sin^4A-\sin^2A}=\dfrac{\sin A}{\cos^2A\cos^3A}\\\\L_s=\dfrac{\cos A(\cos^2A-1)}{\sin A(\sin^2A-1)}+\dfrac{\cos^2A(1-\cos^2A)}{\sin^2A(\sin^2A-1)}\\\\=\dfrac{\cos A(-\sin^2A)}{\sin A(-\cos^2A)}+\dfrac{\cos^2A\sin^2A}{\sin^2A(-\cos^2A)}\\\\=\dfrac{\sin A}{\cos A}-1=\tan A-1

R_s=\dfrac{\sin A}{\cos A\cos^4A}=\dfrac{\sin A}{\cos A}\cdot\dfrac{1}{\cos^4A}=\tan A\cdot\dfrac{1}{\cos^4A}=\dfrac{\tan A}{\cos^4A}


\boxed{L_s\neq R_s}
8 0
3 years ago
The graph below represents Dovans while riding his bike What would be an ordered pair on the line?
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HOPE THIS HELPS
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4 years ago
Adeline sells necklaces for a price, x, and bracelets for a price, y. She expects to sell 35 necklaces and 75 bracelets at a fai
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Answer:

x = 17.046

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