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krek1111 [17]
3 years ago
14

How to find area of a rectangle?

Mathematics
1 answer:
anyanavicka [17]3 years ago
4 0

Formula

Area = L*w


Hint : To find the area of a rectangle , multiply the length times the width


I hope that's help !


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A cylinder has a radius of 6 inches and is 15 inches tall. what is the volume of the cylinder? express the answer in terms of pi
Vikki [24]

The volume of a cylinder that has a radius of 6 inches and is 15 inches tall is 540 π cubic inches

<h3>How to calculate the volume of a cylinder?</h3>

The formula for calculating the volume of cylinder is expressed as:

V = πr²h

where

r is the radius

h isthe height

Given the following parameters

h = 15in

r = 6in

substitute

V =  6^2 * 15π
V  = 540 π cubic inches

Hence the volume of a cylinder that has a radius of 6 inches and is 15 inches tall is 540 π cubic inches

Learn more on volume of a cylinder here: brainly.com/question/9554871

6 0
2 years ago
Please help 40 POINTS
Kamila [148]

Answer:

2.66l

Step-by-step explanation:

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6 0
2 years ago
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Write an equation that represents the line. (-3,-6) and (2,-2) use exact numbers
Ilia_Sergeevich [38]

Given:

The line passes through (-3,-6) and (2,-2).

To find:

The equation of line.

Solution:

If a line passes through two points (x_1,y_1)\text{ and }(x_2,y_2), then the equation of line is

y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

The line passes through (-3,-6) and (2,-2). So, the equation of line is

y-(-6)=\dfrac{-2-(-6)}{2-(-3)}(x-(-3))

y+6=\dfrac{-2+6}{2+3}(x+3)

y+6=\dfrac{4}{5}(x+3)

y+6=\dfrac{4}{5}(x)+\dfrac{4}{5}(3)

Subtract 6 from both sides.

y=\dfrac{4}{5}(x)+\dfrac{12}{5}-6

y=\dfrac{4}{5}(x)+\dfrac{12-30}{5}

y=\dfrac{4}{5}(x)+\dfrac{-18}{5}

y=\dfrac{4}{5}(x)-\dfrac{18}{5}

Therefore, the equation of line is y=\dfrac{4}{5}(x)-\dfrac{18}{5}.

3 0
3 years ago
Answer true or false for 1-10 (picture above)
Softa [21]

\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}\\\\\sqrt{a-b}\geq\sqrt{a}-\sqrt{b}\\----------------------\\1)\ \sqrt{72}=\sqrt{9\cdot8}=\sqrt{9}\cdot\sqrt{8}\qquad TRUE\\\\2)\ \sqrt{13-5}=\sqrt{13}-\sqrt5\qquad FALSE\\\\3)\ \sqrt{9\cdot10}=\sqrt9+\sqrt{10}\qquad FALSE\\\\4)\ \sqrt{18bx}=\sqrt{18}\cdot\sqrt{b}\cdot\sqrt{x}\qquad TRUE\ if\ b\geq0\ and\ x\geq0\\\\5)\ \sqrt{8\cdot12}=\sqrt8\cdot\sqrt{12}\qquad TRUE\\\\6)\ \sqrt{m+x}=\sqrt{m}+\sqrt{x}\qquad FALSE

7)\ \sqrt{100+64}=\sqrt{100}+\sqrt{64}\qquad FALSE\\\\8)\ \sqrt{30}=\sqrt{5\cdot6}=\sqrt5\cdot\sqrt6\qquad TRUE\\\\9)\ \sqrt{44}=\sqrt{22+22}=\sqrt{22}+\sqrt{22}\qquad FALSE\\\\10)\ \sqrt{10\cdot10}=\sqrt{10}\cdot\sqrt{10}\qquad TRUE

3 0
3 years ago
In STU, SU =18 and TU =13. Find m
erastova [34]

Answer:

m=22.2

Step-by-step explanation:

By using the Pythagorean’s theorem i.e

hypotenuse^2=opposite^2+adjacent^2

Where

Hypotenuse =unknown

Opposite =18

Adjacent =13

Hyp^2=18^2+13^2

Hyp^2=324+169

Hyp^2=493

Hyp=sqrt 493

Hyp=m=22.2

3 0
3 years ago
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