Ask question

Ask question

All categories

3 years ago

10

Plz help me find the answers plz i love yu guys so so much??

rmula1" title="20 \div 2" alt="20 \div 2" align="absmiddle" class="latex-formula">

1 answer:

max2010maxim [7]3 years ago

3 0

20 divided by 2 is 10 if that's what your asking.

You might be interested in

The ratio of 6th graders to 7th graders in a MATHCOUNTS club is 5:4. The ratio of 8th graders to 7th graders in the same club is

lorasvet [3.4K]

ANSWER
========

Ratio of 8th graders to 7th graders = 3 : 2

Let the number of graders be 3x and 2x

Number of 8th graders = 18

So , 3x = 18

x = 18/3 = 6

Number of 7th graders = 2 × 6 = 12

Ratio of 6th graders to 7th graders = 5 : 4

Let the number of graders be 5a and 4a

Number of 7th graders = 12

So , 4a = 12

a = 12/4 = 3

Number of 6th graders = 5×3 = 15 ANS

_______________________________

#BE BRAINLY

3 0

3 years ago

Read 2 more answers

Brainlyest fast please

puteri [66]

Answer:

Pleaseee post the question clearly

5 0

3 years ago

The sum of the interior angles of a polygon is 1620 how many sides does the polygon have

Olenka [21]

Answer:

11

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Examples of properties communative, associative, and identity?

BlackZzzverrR [31]

Answer:

Communative: 12 x 5= 5 x 12 Associative: (5) + 6 = (8) + 3 Identity: 5 x 1= 5

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f

WITCHER [35]

Answer:

Part 1) $A=36(\pi-2)\ cm^2$

Part 2) $P=6(\pi+2\sqrt{2})\ cm$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

$A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)$

we have that the base and the height of triangle is equal to the radius of the circle

$r=12\ cm\\b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2$

simplify

Factor 36

$A=36(\pi-2)\ cm^2$

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

$C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r$

substitute the given values

$C=\frac{1}{2}\pi (12)\\C=6\pi\ cm$

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

$AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

Find the perimeter

$P=(6\pi+12\sqrt{2})\ cm$

simplify

Factor 6

$P=6(\pi+2\sqrt{2})\ cm$

4 0

4 years ago