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3 years ago

Scale Drawings question

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

138 is the answer

Step-by-step explanation:

In order to find the perimeter, we use this formula

P=2(l+w)

And to find the width, we use this formula

w = P/2 - L = 18/2 - 3 = 6

So the width is 6

3 × 6 = 18

So thats why the perimeter of the top triangle is 18

We do 21 ÷ 3 = 7, 7 = scale factor

To find the perimeter of the bottom triangle, we do

the width of the top rectangle times 7

so 6 × 7 = 48

So the perimeter for the bottom triangle is

P=2(l+w)=2·(48+21)=138

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The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm.

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Error in the sphere's surface: 29 $cm^2$ and relative error in surface measure: 0.011

Error in the sphere's volume: 205 $cm^3$ and relative error in the volume measure: 0.017

Step-by-step explanation:

(a)

The measured length (l) of the circumference is 90 cm with an error of 0.5 cm, that is:

$l=2\,\pi\,R=90\,cm\\R=\frac{90}{2\,\pi} \,cm=\frac{45}{\pi} \,cm=14.3239\,\,cm$

and with regards to the error:

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then when we use the formula for the sphere's surface, we get:

$S=4\,\pi\,R^2\\dS=4\,\pi\,2\,R\,(dR)\\dS=8\,\,\pi\.(\frac{45}{\pi} \,\,cm)\,(\frac{1}{4\pi}\,cm) =\frac{90}{\pi} \,\,cm^2\approx \,29\,cm^2$

Then the relative error in the surface is:

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(b)

Use the formula for the volume of the sphere:

$V=\frac{4\,\pi}{3} R^3\\dV=\frac{4\,\pi}{3}\,3\,R^2\,(dR)=4\,\pi\,R^2\,(\frac{1}{4\pi}) \,cm=(\frac{45}{\pi})^2 \,\,cm^3\approx 205\,\,cm^3$

Then the relative error in the volume is:

$\frac{dV}{V} =\frac{205}{12310.5} \approx 0.017$

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3 years ago

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TiliK225 [7]

Answer:

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Step-by-step explanation:

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