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4 years ago

From their location in the diagram, what are two possible values for m and n?

Mathematics

1 answer:

Rainbow [258]4 years ago

3 0

Answer:

m =6/2, n = π/2

m = 6/5, n =√2

Step-by-step explanation:

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Roger ran eight laps around a mile track during PE on Monday. Answer each question in the space provided. Part A How many feet d

Gre4nikov [31]

Answer:

A) 10,560 ft. B) 32 more laps.

Step-by-step explanation:

A) Four laps around a track is equal to one mile, and there are 5,280 feet in one mile. If Roger ran eight laps, he ran two miles. Therefore: 5,280 ft. times 2, is equal to 10,560 ft.

B) We know that Roger ran two miles Monday. We also know that four laps around a track is equal to one mile. In order to run 10 miles, he would have to run 40 laps, and since he already ran 8, we can subtract that from 40. You are left with 32 more laps.

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3 years ago

he measurements of two sides of a triangle are 4 cm and 20 cm. What could be a measurement of the third side of the triangle?

hodyreva [135]

Answer:

Step-by-step explanation:

every triangle adds up to 180 sort of, in this case, turn 4 into 40. 40+20=60 so add 80 to be 180!!!

4 0

2 years ago

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:

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