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choli [55]
3 years ago
14

How many 5/8s are in 3?

Mathematics
1 answer:
stira [4]3 years ago
4 0
I believe it to be 4 because 3=24/8

So 5/8 + 5/8 = 10/8 + 5/8 = 15/8 + 5/8 = 20/8 + 5/8 = 25/8, which is the closest number to 3
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The ratio of the circumference of two circles is 3:2.What is the ratio of their area
Nezavi [6.7K]

The ratio of their area is 9/4

<h2>Explanation:</h2>

We use ratios to compares values. In this exercise, we are comparing the circumference of two circles, which is:

3:2

and we want to know what is the ratio of their area. Recall that the circumference of a circle is given by:

C=2\pi r \\ \\ \\ Where: \\ \\ C:Circumference \\ \\ r:Radius \ of \ the \ circle

If we define:

C_{1}: \ Circumference \ of \ circle \ 1 \\ \\ C_{2}: \ Circumference \ of \ circle \ 2 \\ \\ r_{1}: \ Radius \ of \ circle \ 1 \\ \\ r_{2}: \ Radius \ of \ circle \ 2

Then, the ratio of the circumference of two circles is 3:2 is:

\frac{C_{1}}{C_{2}}=\frac{2\pi r_{1}}{2\pi r_{2}}=\frac{3}{2} \\ \\ \therefore \frac{r_{1}}{r_{2}}=\frac{3}{2}

The area of a circle is given by:

A=\pi r^2 \\ \\ A:Area \\ \\ r:Radius

So the ratio of their area can be found as:

\frac{A_{1}}{A_{2}}=\frac{\pi r_{1}^2}{\pi r_{2}^2} \\ \\ \\ A_{1}:Area \ of \ circle \ 1 \\ \\ A_{2}:Area \ of \ circle \ 2

So:

\frac{A_{1}}{A_{2}}=\frac{r_{1}^2}{r_{2}^2}=\left( \frac{r_{1}}{r_{2}} \right)^2 \\ \\ \frac{A_{1}}{A_{2}}=\left(\frac{3}{2}\right)^2 \\ \\ \boxed{\frac{A_{1}}{A_{2}}=\frac{9}{4}}

<h2>Learn more:</h2>

Unit rate: brainly.com/question/13771948#

#LearnWithBrainly

5 0
3 years ago
Oml please don’t say anything that doesn’t help,
Veseljchak [2.6K]

Answer:

first convert mixed fraction to improper fraction then solve by taking LCM

hope the above process helps

6 0
2 years ago
Read 2 more answers
A cone has a volume of 100<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" clas
Bumek [7]

Answer:

12 units

Step-by-step explanation:

The formula for the volume of a cone is given as:

πr²h/3

The volume = 100π cubic units

Diameter = 10 units.

Radius = Diameter/2

= 10 units/2 = 5 units

Hence:

Formula to find the height of a cone

h = 3V/πr²

We substitute

h = 3 × 100π/π × 5²

h = 300π/25π

h = 12 units

The height = 12 units

4 0
3 years ago
PLEASE HELP! THANK YOU! Hint: It’s not C.
Jlenok [28]
The correct answer is A. Each of the y values are 4 times their respective x values
6 0
3 years ago
100 POINTS + BRAINLIST
mojhsa [17]

Problem 1

Since point P is the tangent point, this means angle OPT is a right angle

angle OPT = 90 degrees

Let's use the Pythagorean theorem to find the missing side 'a'

a^2 + b^2 = c^2

a^2 + 7^2 = 12^2

a^2 + 49 = 144

a^2 = 144 - 49

a^2 = 95

a = sqrt(95)

a = 9.7467943

a = 9.7

----------------

Let's use the sine and arcsine rule to find angle y

sin(angle) = opposite/hypotenuse

sin(y) = 7/12

y = arcsin( 7/12 )

y = 35.6853347

y = 35.7

The value of y is approximate. Make sure your calculator is in degree mode. Arcsine is the same as inverse sine or \sin^{-1}

====================================================

Problem 2

Focus on triangle OHP. This may or may not be a right triangle. The goal is to test if it is or not.

We use the converse of the Pythagorean theorem to check.

Recall that the converse of the Pythagorean theorem says: If a^2+b^2 = c^2 is a true equation, then the triangle is a right triangle. The value of c is always the longest side. The order of a and b doesn't matter.

In this case we have

  • a = 7
  • b = 13
  • c = 16

Which leads to

a^2 + b^2 = c^2

7^2 + 13^2 = 16^2

49 + 169 = 256

218 = 256

We get a false equation at the end, since both sides aren't the same number, which means the original equation is false.

We don't get a^2 + b^2 = c^2 to be true, therefore this triangle is not a right triangle.

Consequently, this means angle OPH cannot possible be 90 degrees (if it was then we'd have a right triangle). Therefore, point P is not a tangent point.

You follow the same basic idea for triangle OQH and show that point Q is not a tangent point either. Or you could use a symmetry argument to note that triangle OPH is a mirror reflection of triangle OQH over the line segment OH. This implies that whatever properties triangle OPH has, then triangle OQH has them as well for the corresponding pieces.

7 0
2 years ago
Read 2 more answers
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