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vova2212 [387]
3 years ago
12

8 out of 200 is what percent

Mathematics
2 answers:
kipiarov [429]3 years ago
6 0
Hey there! 

Here are the steps involved in answering this question:

1. Write 8 out of 200 as a fraction which is \frac{8}{200}

2. Turn the fraction into a decimal. To do this, divide the numerator by the denominator. The numerator is the top part of the fraction while the denominator is the bottom part of the fraction.

\frac{8}{200}  = 0.04

3. You get 0.04. Now, turn this into a percent. Move the decimal, 2 places to the right or just multiply the decimal by 100.

0.04 = 4%

The final answer is 4%.

Another way of doing this is to get 100 as a denominator.
Since, percent means, "Out of 100", we have to make the denominator 100.

1.  \frac{8}{200} = 4/100

Just divide each side by 2.

Again, the final answer is, 4%

<span>(If you feel that my answer has helped you, please consider rating it and giving it a thank you! Also, feel free to make the best answer, the brainliest answer!)
</span>
<span>Thank you! :D</span>
Jobisdone [24]3 years ago
4 0
Divide both by 2 so it becomes on 100
4 and 100
8 out of 200 is 4%
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3 years ago
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
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