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statuscvo [17]
2 years ago
11

There are 40 boys and 16 girls in a class of students. What is the ratio of girls to students?

Mathematics
2 answers:
vichka [17]2 years ago
7 0

Answer:

4: 13

Step-by-step explanation:

every 4 girls there's 13 boys

hope this helps.

have a great day!

HACTEHA [7]2 years ago
6 0

Answer:

Hello! To find the ratio of girls to total students, find the number of students first:

40 + 16 = 56 (The total number of students is 56.)

The initial ratio of girls to total students is 16:56 - out of 56 students, there are 16 girls. To simplify this ratio, divide by the greatest common factor (8):

16:56 / 8 = 2:7

The simplified ratio is 2:7 - out of every 7 students, there are 2 girls.

Hope this helps! Have a great day - Mani :)

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1) 36

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3 * 12 = 36

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What is the code? Thank you! Whoever answers this, I will give brainliest!
shepuryov [24]

Answer:

5821

Step-by-step explanation:

For the first symbol, the number is 5.

In the first problem, the digit corresponding to that symbol is 5.

In the second one, it should be 5. The area of a triangle is base * height / 2. The base is 15 and the height is 9. Therefore, the area will be 15 * 9 / 2 = 67.5 square units. Now, the number corresponding to that symbol is 5.

In the fourth problem, the number corresponding to the symbol is also 5.

For the second symbol, the number is 8.

In the 5th problem, the number corresponding to that symbol is 8.

In the 7th problem, the number corresponding to the symbol is 8.

In the last problem, 96 is the correct answer, and everything is in the right place. I assume this must be the mistake of the maker of the problem.

For the third symbol, the number is 2

In the 1st and 3rd problems, the number corresponding to that symbol is 2

Finally, for the last symbol, the number is 1

In the 4th problem, there is a trapezoid. The formula for the area of a trapezoid is base 1 * base 2 / 2 * height. base 1 = 29, base 2 = 13. 13 + 29 = 42. 42 / 2 = 21. 21 * 15 = 315. Now, the 2nd digit, the one corresponding to the symbol, is 1.

In the last problem, you need to find the surface area of a figure. I would do this by splitting it up. The area for a triangle is base * height / 2. That means, the area of the triangle would be 8 * 7 / 2 = 28. There are four identical triangles, so you multiply this by 4 and get 112. Next, you find the area of the square which is 7 * 7. 7 * 7 = 49. Now, you add them together. 112 + 49 = 161. Now, the first and last digits corresponding to the symbol are both 1.

The answer is 5821.

4 0
3 years ago
Please help with this calculus problem!
ELEN [110]
\dfrac{\mathrm dQ}{\mathrm dt}=\dfrac{a(1-5bt^2)}{(1+bt^2)^4}
Q(t)=\displaystyle\int\frac{a(1-5bt^2)}{(1+bt^2)^4}\,\mathrm dt

Let t=\dfrac1{\sqrt b}\tan u, so that \mathrm dt=\dfrac1{\sqrt b}\sec^2u\,\mathrm du. Then the integral becomes

\displaystyle\frac a{\sqrt b}\int\frac{1-5b\left(\frac1{\sqrt b}\tan u\right)^2}{\left(1+b\left(\frac1{\sqrtb}\tan u\right)^2\right)^4}\sec^2u\,\mathrm du
=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{(1+\tan^2u)^4}\sec^2u\,\mathrm du
=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{\sec^6u}\,\mathrm du
=\displaystyle\frac a{\sqrt b}\int(\cos^6u-5\sin^2u\cos^4u)\,\mathrm du
=\displaystyle\frac a{\sqrt b}\int(6\cos^2u-5)\cos^4u\,\mathrm du

One way to proceed from here is to use the power reduction formula for cosine:

\cos^{2n}x=\left(\dfrac{1+\cos2x}2\right)^n

You'll end up with

=\displaystyle\frac a{16\sqrt b}\int(5\cos2u+8\cos4u+3\cos6u)\,\mathrm du
=\displaystyle\frac a{16\sqrt b}\left(\frac52\sin2u+2\sin4u+\frac12\sin6u\right)+C
=\dfrac a{32\sqrt b}(5\sin2u+4\sin4u+\sin6u)+C
Q(t)=\dfrac{at}{(1+bt^2)^3}+C
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