How much larger is the area of the rectangle than the

Mathematics

1 answer:

lana [24]3 years ago

5 0

Answer:

The area of the rectangle is 8 cm 2. 1. Find the area and perimeter of a rectangle with a length of 6 cm and a breadth of 4 cm. 2. A square table cloth has a side of 5 m.

Step-by-step explanation:

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forty percent of participants in a math contest were boys. fifty percent of the boys revived medals. the number of boys who rece

aleksklad [387]

20% of the participants are girls and got a medal, and 33% of the girls got a medal.

<h3>What percent of the participants were girls who received medal?</h3>

First, we know that 40% of the participants were boys, so the other 60% were girls.

We know that 50% of the boys got a medal, so the percentage (in decimal form) of participants that are boys and got a medal is:

P = (0.4)*(0.5) = 0.2

So, 20% of the participants are boys and got a medal.

We know that the number of girls that got a medal is the same as the number of boys, then we also have that 20% of the participants are women who got a medal.

Then if the percentage of girls that got a medal is x (in decimal form), we must have that:

Q = (0.6)*(x) = 0.2

x = 0.2/0.6 = 0.33

x = 0.33

This means that 33% of the girls got a medal.

If you want to learn more about percentages:

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3 0

2 years ago

Multiply or divide. Show your work.<br><br> 3n²-n/n²-1÷n²/n+1

Ugo [173]

$\frac{3n^{2} - n}{n^{2} - 1} \div \frac{n^{2}}{n + 1}$
$\frac{n(3n) - n(1)}{n^{2} + n - n - 1} \div \frac{n^{2}}{n + 1}$
$\frac{n(3n - 1)}{n(n) + n(1) - 1(n) - 1(1)} \div \frac{n^{2}}{n + 1}$
$\frac{n(3n - 1)}{n(n + 1) - 1(n + 1)} \div \frac{n^{2}}{n + 1}$
$\frac{n(3n - 1)}{(n - 1)(n + 1)} \div \frac{n^{2}}{n + 1}$
$\frac{n(3n - 1)}{(n - 1)(n + 1)} * \frac{n + 1}{n^{2}}$
$\frac{3n - 1}{n - 1} * \frac{1}{n}$
$\frac{3n - 1}{n(n - 1)}$

5 0

3 years ago

Which expression is equivalent to StartFraction c squared minus 4 Over c + 3 EndFraction divided by StartFraction c + 2 Over 3 (

Firdavs [7]

Answer:

Step-by-step explanation:

Given:

$\dfrac{c^2-4}{c+3}\div \dfrac{c+2}{3(c^2-9)}$