20 is what percent of 50?

1 answer:

3 0

We can rephrase the question as "What percent of 50 is 20?".
"What percent ..." is an unknown. We'll use the variable x to represent this unknown.

The word 'of' in mathematics can mean 'multiply'. Likewise the word 'is' can mean 'equal'.

Now we have the phrase "20 equals to x times 50".

Or in an equation: 20 = x * 50

Solve for x.

20 = x * 50
$\frac{20}{50}$ = x <-- Divide both sides by 50

Simplify the fraction.

$\frac{20}{50} = \frac{2}{5}$

Convert the fraction into a decimal.

$\frac{2}{5}$ = 0.4

Convert the decimal into a percent.

0.4 = 40%

So, 40% is the answer.

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Determine and prove what shape is formed for the given coordinates for ABCD, and then find the

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Answer:

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Step-by-step explanation:

Plot points A (10,-4), B (6,-7), C (3,-3), D (7,0) on the coordinate plane.

Find slopes of the sides:

$AB:\ \dfrac{-7-(-4)}{6-10}=\dfrac{3}{4}\\ \\BC:\ \dfrac{-3-(-7)}{3-6}=-\dfrac{4}{3}\\ \\CD:\ \dfrac{0-(-3)}{7-3}=\dfrac{3}{4}\\ \\DA:\ \dfrac{-4-0}{10-7}=-\dfrac{4}{3}$

The slopes of opposite sides are the same, so opposite sides are parallel. The slopes of adjacent sides have product of -1, then adjacent sides are perpendicular.

Find the lengths of all sides:

$AB=\sqrt{(10-6)^2+(-4-(-7))^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\BC=\sqrt{(6-3)^2+(-7-(-3))^2}=\sqrt{3^2+4^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\CD=\sqrt{(3-7)^2+(-3-0)^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\AD=\sqrt{(10-7)^2+(-4-0)^2}=\sqrt{3^2+4^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\$