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Art [367]
2 years ago
9

What is the area of base for 7mm height and 7mm radius cylinder?​

Mathematics
1 answer:
sveta [45]2 years ago
3 0

Answer:

153.94 mm^2

Step-by-step explanation:

area of base = area of a circle = Pi * radius squared

radius squared = 49

area of base =  Pi * 49

area of base =  153.94

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Tools used to physically represent numbers​
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Step-by-step explanation:

8 0
3 years ago
What is the answer to this question?
andrey2020 [161]

Answer:

EF/IH and AD

Step-by-step explanation:

diameter is 2x radius

4 0
3 years ago
I WILL GIVE BARINLEST WHO HELP ME<br><br>3
Pavlova-9 [17]

Answer:

whats the question

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Solve.<br><br> y = –4x + 10<br> y = –3x + 5
saw5 [17]
The answers in order are
Y= 10/4
Y=5/3
8 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

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

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0
3 years ago
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