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4 years ago

Write a real world problem that involves finding the area is two circles

1 answer:

8 0

An architect is trying to build a cover for two cylinders. He knows the radius of Cylinder 1 is 15 inches and the radius of Cylinder 2 is 16 inches. What is the area of both circles?

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What is the absolute value of the complex number -4- sqrt 2i? A) sqrt 14 B) 3sqrt2 C)14 D)18

Novosadov [1.4K]

Remember the pythagorean theorem.

a^2 + b^2 = c^2

The complex number is a+bi, the absolute value is 'c'

$|-4-\sqrt{2} i| = \sqrt{(-4)^2 + (-\sqrt{2})^2} = \sqrt{16+2} = \sqrt{18} = 3\sqrt{2}$

7 0

3 years ago

I need the answer please help

levacccp [35]

Answer:

2k + 5

Step-by-step explanation

did it on e2020

3 0

3 years ago

Simplify 15c3 please

aleksklad [387]

Answer:

455.

Step-by-step explanation:

Yes. You require the number of combinations of 3 from 15.

15C3

The formula for this is 15! / 12! 3! or you can do itr as follows

15C3 = 15*14*13/3*2*1

= 5*7*13

= 35*13

= 455.

7 0

3 years ago

Help me and i’ll mark you brainliest plus you get 20 points !!!!

Anton [14]

Answer:

Angle bisector

Step-by-step explanation:

Find the measure of the angle COA. By angle addition postulate,

$m\angle COX=m\angle AOX+m\angle COA$

From the diagram,

$m\angle COX=80^{\circ}\\ \\m\angle AOX=40^{\circ},$

then

$80^{\circ}=m\angle COA+40^{\circ}\\ \\m\angle COA=80^{\circ}-40^{\circ}=40^{\circ}$

Find the measure of the angle BOA. By angle addition postulate,

$m\angle BOX=m\angle AOX+m\angle BOA$

From the diagram,

$m\angle BOX=60^{\circ}\\ \\m\angle AOX=40^{\circ},$

then

$60^{\circ}=m\angle BOA+40^{\circ}\\ \\m\angle BOA=60^{\circ}-40^{\circ}=20^{\circ}$

Find the measure of the angle COB. By angle addition postulate,

$m\angle COX=m\angle BOX+m\angle COB$

From the diagram,

$m\angle BOX=60^{\circ}\\ \\m\angle COX=80^{\circ},$

then

$80^{\circ}=m\angle COB+60^{\circ}\\ \\m\angle COB=80^{\circ}-60^{\circ}=20^{\circ}$

This means, the measures of angles COB and BOA are the same and are equal half the measure of angle COA, so angles COB and BOA are congruent. This means, the ray OB is the angle bisector of angle COA

6 0

3 years ago

The table shows the number of world-wide subscribers, in millions, each quarter from 2012 to 2017 for a company that provides

Rufina [12.5K]

Please find full question attached

Answer:

Option B is correct

Step-by-step explanation:

We have made a sketch of the line chart using the data provided in the question. From the attachment and the option selected it is evident that as the year goes by the number of subscribers increases. In other words the line chart shows that subscribers increased consistently from 2006 to 2011

4 0

3 years ago