What number is 15% of 80? use percent proportions

1 answer:

8 0

!5% is equivalent to 15 parts over the whole 100 part. It is written in proportions as 15/100.

Similarly, you need to find how many parts in 80 are equivalent to 15%or 15/100. So, just equate both equation with x as an unknown.

x/80 = 15/100
x = (15/100)*80
x = 12

Thus, the 15% of 80 is 12.

You might be interested in

64/100 as a simplified fraction

spin [16.1K]

64/100
= (64/4) / (100/4)
= 16/25

The final answer is 16/25~

6 0

3 years ago

Read 2 more answers

BRAINLIEST!!!!!!!

Murrr4er [49]

All the angles inside a quadrilateral add to 360 degrees.
But first, we should solve for x
Opposite angles add up to 180 degrees.
So we'll use angles B and D
x + 10 + x + 24 = 180
combine like terms
2x + 34 = 180
subtract 34 from both sides
2x = 146
divide both sides by 2
x = 73

Now let's solve for the angles
Angle D: x + 24 = 97
Angle B: x + 10 = 83
Angle A: x + 15 = 88

Opposite angles add up to 180
we know that angle A is 88
C + 88 = 180
subtract 88 from both sides
C = 92

Hope this helps!

4 0

3 years ago

Please answer this<br><br> (I don’t know what else to put, it need 20 characters)

Vinvika [58]

Answer:

15yd^2

Step-by-step explanation:

Triangle area formula is A = bh/2

A = 6*5/2

A = 30/2

A = 15

Best of Luck!

6 0

3 years ago

Read 2 more answers

Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos

Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$