7. {MCC.7.G.3} Name the cross sectional area below. *

Mathematics

1 answer:

Rzqust [24]3 years ago

5 0

It’s an isosceles Triangle.

Looking at the picture it seems that both slant height are the same.

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I have a roof that measures 175 ft by 45 ft. During the hurricane 11 inches of rain fell.

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Answer:

4.1 billion

Step-by-step explanation:

1 ft = 30.48 cm

1 in = 2.54 cm

The volume of rain that fell on the roof is given by ...

V = LWH

V = (175 ft × 30.48 cm/ft)(45 ft × 30.48 cm/ft)(11 in × 2.54 cm/in)

= 175×45×11×30.48²×2.54 cm³ = 204,412,236.336 cm³

At 20 drops per cm³, this will be ...

20×204,412,236.336 ≈ 4,088,244,727 . . . . raindrops

About 4.1 billion raindrops fell on your roof.

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Can I get help answering this question? It's a derivatives question, further info in the attachment.

Nadusha1986 [10]

$\bf f(x)=|3+9x|\implies f(x)=\sqrt{(3+9x)^2}\implies f(x)=[(3+9x)^2]^{\frac{1}{2}}
\\\\\\
\cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{2}[(3+9x)^2]^{-\frac{1}{2}}\cdot 2(3+9x)\cdot 9}\implies \cfrac{dy}{dx}=\cfrac{9(3+9x)}{[(3+9x)^2]^{\frac{1}{2}}}
\\\\\\
\cfrac{dy}{dx}=\cfrac{27+81x}{|3+9x|}\\\\
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$\bf \textit{left-hand derivative at }x=-\frac{3}{9}\implies \cfrac{27+81x}{-(3+9x)}
\\\\\\
\cfrac{27+81\left( -\frac{3}{9} \right)}{-3-9\left( -\frac{3}{9} \right)}\implies \cfrac{27-27}{-3+3}\implies \stackrel{unde fined}{\cfrac{0}{0}}\\\\
-------------------------------\\\\
\textit{right-hand derivative at }x=-\frac{3}{9}\implies \cfrac{27+81x}{+(3+9x)}
\\\\\\
\cfrac{27+81\left( -\frac{3}{9} \right)}{3+9\left( -\frac{3}{9} \right)}\implies \cfrac{27-27}{3-3}\implies \stackrel{unde fined}{\cfrac{0}{0}}$

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