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Dennis_Churaev [7]
2 years ago
11

Jasmine’s race time was 34.287 minutes.round her race time to the nearest 10th of a minute

Mathematics
2 answers:
amm18122 years ago
7 0
Hi!

34.287 rounded to the nearest tenth is 34.290

This is because 7 is greater than 5, so we would round it to the cloest tenth that is slightly higher than 7. =)
stich3 [128]2 years ago
3 0
34.3 minutes because if a number is higher than 5 it rounds up and in this case since the decimals are 2 tenths 8 hundredths and 7 thousandths. because the hundredths value is a number bigger than 5 you would round it up to 1 tenth making the tenths value 3
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You use a line of best fit for a set of data to make a prediction about an unknown value. the correlation coeffecient is -0.833
alina1380 [7]

Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2  = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2  = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

5 0
2 years ago
I need to find the volume of a nose cone.
rjkz [21]

well then, the volume of the nose cone will just be the sum of the volume of the cylinder below and the cone above.

since the diameter for both is 8, then their radius is half that, or 4.

\bf \stackrel{\textit{volume of a cone}}{V=\cfrac{\pi r^2 h}{3}}~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} r=4\\ h=6 \end{cases}\implies V=\cfrac{\pi (4)^2(6)}{3}\implies V=32\pi \\\\\\ \stackrel{\textit{volume of a cylinder}}{V=\pi r^2 h}~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} r=4\\ h=6 \end{cases}\implies V=\pi (4)^2(6)\implies V=96\pi \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{volume of the nose cone}}{32\pi +96\pi \implies 128\pi }\qquad \approx \qquad 402.12

8 0
3 years ago
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Blababa [14]

Answer:

1/10

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Step-by-step explanation:

7 0
3 years ago
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dedylja [7]
We shall use a negative sign for an elevation below sea level, and a positive sign for an elevation above sea level.

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Part A.
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Part B.
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Answer:
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xeze [42]
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