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3 years ago

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A young sumo wrestler decided to go on a special high-protein diet to gain weight rapidly and at a constant rate. After 8 88 mon

ths, he weighed 138 138138 kilograms. He started at 90 9090 kilograms. Let y yy represent the sumo wrestler's weight (in kilograms) after x xx months. Complete the equation for the relationship between the weight and number of months.

1 answer:

Aliun [14]3 years ago

3 0

Answer:

Equation for the relationship between the weight and number of months -

$W(t) = W(0) + 6 * t$

Step-by-step explanation:

Given

The initial weight of young sumo wrestler just before he decide to gain more weight is equal to $90$ kilo grams.

The new weight of young sumo wrestler after eight months is equal to $138$ kilo grams.

Weight gained by sumo wrestler during the period of eight months

$138 -90\\= 48$

The rate at which weight was gained

$\frac{48}{8}$

$6$ kilo grams per month

Let W(t) depict the weight after 8 months

t depict the time period in moths

So the equation would be

$W(t) = W(0) + 6 * t$

Where

$W(0) =$ Initial weight before 8 months

$t =$ time in months

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4 years ago

a jogger runs 4 miles per hour faster downhill than uphill. if the jogger can run 5 miles downhill in the same time that it take

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Answer: the rate uphill is 6 mph.

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Let x represent the rate at which the jogger ran uphill.

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4 years ago

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4 years ago

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The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. A) Use differentials to estimate the max

siniylev [52]

Answer:

A) The maximum error in the calculated surface area: $25cm^2$

Relative error: $0.013$

B) The maximum error in the calculated volume: $162cm^2$

Relative error: $0.019$

Step-by-step explanation:

A) The formula for the surface area is:

$A=4\pi r^2$

The measured value is the circumference which is equal to:

$C=2\pi r$

then the radius is:

$r=\frac{C}{2\pi}$

Substituting in the formula of the surface:

$A=4\pi(\frac{C}{2\pi})^2\\A=4\pi(\frac{C^2}{4\pi^2})\\A=\frac{C^2}{\pi}$

Using the formula to calculate the error:

$dy=f'(x)dx$

Where $x$ is the variable measured and $y$ is a function of $x$ ( $y=f(x)$ ).

$dA=f'(C)dC\\dA=\frac{2C^{(2-1)}}{\pi}dC\\dA=\frac{2C}{\pi}dC$

We have C=80cm and dC=0.5cm

$dA=\frac{2C}{\pi}dC\\dA=\frac{2(80)}{\pi}(0.5)\\dA=\frac{160}{\pi}(0.5)\\dA=50.9296(0.5)\\dA=25.4648\approx25cm^2$

The relative error is the maximum error divide by the total area. The total area is: $A=\frac{C^2}{\pi}=\frac{(80)^2}{\pi}=\frac{6400}{\pi}=2037.1833cm^2$

$\frac{dA}{A}=\frac{25.4648}{2037.1833} =0.0125\approx0.013$

B) The formula for the volume is:

$V=\frac{4}{3} \pi r^3$

Using $r=\frac{C}{2\pi}$

$V=\frac{4}{3} \pi r^3\\V=\frac{4}{3} \pi (\frac{C}{2\pi})^3\\V=\frac{4}{3} \pi (\frac{C^3}{8\pi^3})\\V=\frac{1}{3}(\frac{C^3}{2\pi^2})\\V=\frac{C^3}{6\pi^2}$

The maximum error is:

$dV=\frac{3C^{3-1}}{6\pi^2}dC\\dV=\frac{C^{2}}{2\pi^2}dC\\dV=\frac{(80)^{2}}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=(324.2278)(0.5)\\dV=162.1139\approx162cm^2$

The calculated volume is:

$V=\frac{C^3}{6\pi^2}\\V=\frac{(80)^3}{6\pi^2}\\V=\frac{512000}{6\pi^2}\\V=8646.0743$

The relative error is:

$\frac{dV}{V}=\frac{162.1139}{8646.0743}=0.0188\approx0.019$

3 0

3 years ago