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Akimi4 [234]
2 years ago
8

The distance from the center of a circulartrashcan to the edge of the trashcan is 15 inches.What is the area of the top of the t

rashcan?
Mathematics
1 answer:
Black_prince [1.1K]2 years ago
5 0

Answer:

706.95in²

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, a cylinder

Step one

The trash can has a cylindrical shape

The distance from the center to the edge is the radius r= 15in

Step two

Area of top (circular ) = πr²

A= 3.142*15²

A= 3.142*225

A= 706.95in²

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7 * 2.9

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Consider a circle centered at C(-2,-4). If the point P(1,-1) lies on the circle, then which of the following points also lies on
Aliun [14]

Answer:

Consider a circle centered at C(-2,-4). If the point P(1,-1) lies on the circle, then which of the following points also lies on the circle?

<h3>A. (-2, -√18)</h3>

B. (-2+√18, -4)

C. (√18, -4)

D. (-2, 4+√18)

Step-by-step explanation:

  • If the distance is greater than the radius, the point lies outside. If it's equal to the radius, the point lies on the circle. And if it's less than the radius, you guessed it right, the point will lie inside the circle.

hope it's help you

7 0
2 years ago
Can you please help me thank you so much
77julia77 [94]

Answer:

I would say angle 6 is supplementary to angle 3.

Step-by-step explanation:

Supplementary angles are angles that add up to 180 degrees.

Angles 6 and 3 are right angles.

90 + 90 = 180

Hope I helped!

8 0
2 years ago
Read 2 more answers
The following data set represents the age of
Julli [10]

\huge\text{Hey there!}


\huge\textsf{Breaking the meaning down in simpler terms}
\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}


\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}

\huge\textsf{Formulas for each term}

\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }


\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}


\huge\textbf{SOLVING FOR THE QUESTION}

\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of  \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}

\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}

\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}


\huge\text{Good luck on your assignment \& enjoy your day!}

<h3>
~\frak{Amphitrite1040:)}</h3>
8 0
2 years ago
Can someone help, please?
Sunny_sXe [5.5K]
You have to first mess around with the first shape, ABCD, and split that into a rectangle and a right triangle. once you do that, it's pretty painstaking, but simple.

if you look at it you can tell that EFGH is just half the size, but the same ratios and everything.

So, you would just take every perimeter measurement from ABCD, and divide it by two and then sum them together.

2.5 + 1.5 + 4.0 + 2.0 = 10
6 0
2 years ago
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