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SVETLANKA909090 [29]
3 years ago
9

Using y = 1 − 2x, plot the ordered pairs from the table. Then graph the function represented by the ordered pairs and tell wheth

er the function is linear or nonlinear.
Helppp plzz due in 20 min

Mathematics
1 answer:
svetlana [45]3 years ago
7 0
I think therefore its a linear function

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c

Step-by-step explanation:

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The circumference of a sphere was measured to be 90 cm with a possible error of 0.5 cm.
Akimi4 [234]

Answer:

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:

dl=0.5 \, cm\\dl=2\,\pi\,dR\\dR=\frac{dl}{2\,\pi} =\frac{1}{4\,\pi} cm = 0.0796\,cm

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:

\frac{dS}{S} =\frac{90/\pi}{4\,\pi\,R^2} =\frac{1}{90} =0.011

(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

3 0
3 years ago
Solve each system by elimnination.<br> 5c- 4t= 8<br> 14 + 4t = 3c
VladimirAG [237]
The answer is c=11 and t=11.75 :)

Here’s the explanation of why and why it is correct

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3 years ago
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