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maxonik [38]
2 years ago
11

Find the volume of the cylinder. Round your answer to the nearest tenth. 16 cm 8 cm The volume of the cylinder is about (blank)

cubic centimeters.​

Mathematics
1 answer:
attashe74 [19]2 years ago
7 0

Answer:

please mark me brainliest if it helps you

804.5cm^3

800 cm^3

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Find BC if AC = 13 and AB = 10
maxonik [38]

Answer:

BC=3

Step-by-step explanation:

<em>We </em><em>can </em><em>solve </em><em>this </em><em>by </em><em>eliminating</em><em> </em><em>given </em><em>detail </em><em>of </em><em>A.</em>

AC=13 , AB=10

<u>To </u><u>find </u><u>BC,</u><u> </u><u>We </u><u>minus </u><u>A</u><u>C</u><u> </u><u>With </u><u>AB</u>

BC= AC-AB

<u>BC=3</u>

3 Is the final answer

I hope this helps, dont hesitate to ask for any question.

Mark me as brainliest is appreciated.Tq!!

7 0
3 years ago
The blue segment below is a diameter of O. What is the length of the radius<br>of the circle?​
Black_prince [1.1K]

Answer:

the radius is half of O

Step-by-step explanation:

5 0
3 years ago
Simply<br> tan 45°. sin 30° - cot 45°/sec 60°​
dusya [7]

If we want to write the given four numbers in another form, we can write it like this;

tan45=sin45/cos45=1

sin30=sin(\frac{\pi }{2}-60 )=cos60=\frac{1}{2}

cot45=cos45/sin45=1

sec60=1/cos60=\frac{1}{1/2} =2

Now let's rewrite the given expression and get the result.

(tan45.sin30)-(cot45/sec60)=(1.\frac{1}{2})-(\frac{1}{2} )=0

5 0
1 year ago
Read 2 more answers
Question below here you can rear if not aks me please and thank you ​
Vanyuwa [196]
180-48.2-75=56.8 thank you
5 0
2 years ago
Question 18
Stels [109]

The system of linear equations that best represents the situation are:

z+y=18

3z +2y=44

<h3>What are the linear equations?</h3>

In order to determine the values of the two types of cheese, two set of linear equations can be formed from the question. The equations have to be solved together using either the elimination , substitution or graphical methods.

To learn more about simultaneous equations, please check: brainly.com/question/25875552

#SPJ1

8 0
1 year ago
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