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aalyn [17]
3 years ago
11

15% of Alina’s workout consisted of cardio exercise. 25% of Dane’s workout consisted of cardio exercise. Which statement must be

true?
If they worked out for the same amount of time, Dane did 10 more minutes of cardio exercise than Alina.


If Alina’s workout was 80 minutes, and Dane’s workout was 48 minutes, they spent the same amount of time on cardio exercise.


If both workouts were 60 minutes, Dane spent 8 more minutes on cardio exercise than Alina did.


Alina spent less time on cardio exercise than Dane did.
Mathematics
1 answer:
Arisa [49]3 years ago
6 0


The correct answer is B. If Alina’s workout was 80 minutes, and Dane’s workout was 48 minutes, they spent the same amount of time on cardio exercise.

To figure this out, first find 15% of 80 and then 25% of 48.

To find 15% of 80, you can first find 10% of 80 which is 8(just move the decimal one place to the left). The remaining 5% will be half of 8 since 8 is 10%. Add both the numbers up and you will get 12. (8+4=12)

To find 25% of 48, you can divide 48 by 4. Since percent means 'out of 100', 25% is 25 of 100. 100 divided by 25 is equal to 4. So, 48 divided by 4 is equal to 12.

So, the final answer is B. They both equal to 12.

Hope it helps :)

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Naddik [55]
Just a guess, but the angle inside the triangle that's not x or y is the remaining angle after subtracting the 30 degrees shown.  That makes the non labeled angle inside the triangle 150 degrees.  since all triangles are 180 degrees, this means x and y add up to 30.  They look roughly the same, so i'd guess they're both 15 degrees.
6 0
3 years ago
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
Find the value of x.
Travka [436]

Answer:X=9

Step-by-step explanation:

x+18=3x

18=2x

x=9

4 0
3 years ago
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love history [14]

Answer:

m = - 7

Step-by-step explanation:

Given

6(m - 1) = 3(3m + 5) ← distribute parenthesis on both sides

6m - 6 = 9m + 15 ( subtract 9m from both sides )

- 3m - 6 = 15 ( add 6 to both sides )

- 3m = 21 ( divide both sides by - 3 )

m = - 7

8 0
3 years ago
Ok pls help ***6TH GRADE*** its 50 points since its due in 15 I'm desperateeeeeee. i HAVE ASKED SO MANY TIMES AND NO ONE HELPS S
Harlamova29_29 [7]

Answer:

For the first pic, line two is 8 km

I don't know about the second one. This is the best I can do, sorry

I hope I helped somehow, have a good day

8 0
3 years ago
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