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Ipatiy [6.2K]
3 years ago
11

dennis ran a mile in 593.7 seconds.martina ran a mile in 573.36 seconds. what was the difference in their running times? A-5.14

s B-6.01 C - 20.34 s D -26.01 s
Mathematics
2 answers:
Svetlanka [38]3 years ago
8 0

subtract them

593.7-573.36 = 20.34seconds

 so c is the answer


harina [27]3 years ago
4 0
C- subtract Martina's from Dennis's.
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(i) Given that

V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

Compute the integral:

V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr

V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R

V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)

V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

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3 years ago
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3 years ago
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How much more does it need to, get to 6,550,000?


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3 years ago
Given EG = 16 and FH = 12, what is the length of one side of the rhombus? 6 units 8 units 10 units 14 units PLEASE HELP ONLY COM
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EG and FH are diagonals of the rhombus and the bisect each other at the centre to for a righ angle triangle with the side of the rhombus as the hypothenus.
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Answer:

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