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riadik2000 [5.3K]
1 year ago
6

Date:

Mathematics
1 answer:
katrin2010 [14]1 year ago
5 0

The volume of any shape is the amount of space in it.

The volume of the cylinder is 1570 square units

<h3>How to determine the volume?</h3>

The question is incomplete. Assume the following parameters

Shape = Cylinder

Radius, r = 10

Height, h = 5.

The volume of a cylinder is:

V = πr^2h

So, we have:

V = 3.14 * 10^2 * 5

Evaluate the product

V = 1570

Using the assumed values, the volume of the cylinder is 1570 square units

Read more about volumes at:

brainly.com/question/1972490

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HI PLEASE ANSWER THIS QUESTION THANK YOU SO MUCH!
krek1111 [17]

1.  m³

2. cm³

3. s³

<h3>What is Unit?</h3>

A unit of measurement is a definite magnitude of a quantity, defined and adopted by convention or by law, that is used as a standard for measurement of the same kind of quantity.

1. Volume is measure in m³

2.  volume of a rectangular pencil case in cm³

3. V = s x s x s or s³

Learn more about volume here:

brainly.com/question/1324203

#SPJ1

3 0
1 year ago
Solve log x=2 by changing it to exponential form.
Anton [14]
Here is a video of how to do it! https://www.youtube.com/watch?v=c-FVjcPpz54
8 0
3 years ago
Find cos B exactly if a = 15, b = 11, and angle Cis a right angle
astra-53 [7]

Answer:

Final answer is \cos\left(B\right)=\frac{15}{\sqrt{346}} which is the last choice.

Step-by-step explanation:

In triangle ABC, we have been given a = 15, b = 11, and angle C is a right angle.

Now using those values, we need to find the exact value of cos B.

Apply Pythagorean formula:

AB^2=AC^2+BC^2

AB^2=11^2+15^2

AB^2=121+225

AB^2=346

AB=\sqrt{346}

Now apply formula of cos

\cos\left(B\right)=\frac{BC}{AB}

\cos\left(B\right)=\frac{15}{\sqrt{346}}

Hence final answer is \cos\left(B\right)=\frac{15}{\sqrt{346}}


3 0
3 years ago
Quadrilateral ABCDABCD has vertices AA (-2, 2), BB (4, 5), CC (3, 0) and DD (-3, -3). What are the coordinates of the midpoint o
Arada [10]
Check the picture below.

the quadrilateral is a parallelogram, thus the diagonals bisect each other, so, we could have used either diagonal, and get the same midpoint.


\bf \textit{middle point of 2 points }\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;B&({{ 4}}\quad ,&{{ 5}})\quad &#10;%  (c,d)&#10;D&({{ -3}}\quad ,&{{ -3}})&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)&#10;\\\\\\&#10;\left(\cfrac{{{ -3+4}} }{2}\quad ,\quad \cfrac{-3+5}{2} \right)

6 0
3 years ago
What is the shape callled that looks like a rectangle with two sides that look like a triangle
Softa [21]
Quadrangle
A quadrilateral can sometimes be called: a Quadrangle ("four angles"), so it sounds like "triangle"
Please Mark Brainliest.
8 0
2 years ago
Read 2 more answers
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