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Lelu [443]
3 years ago
14

Suppose $10000 is deposited in a bank at 3.5% annual interest. what is the interest earned after one year

Mathematics
1 answer:
alexgriva [62]3 years ago
8 0
I=PRT\implies I=\$10000\times0.035\times1=\$350
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What is the area of a bedroom thats 12 ft in length and 10 ft in width?
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Answer:

120 sq ft.

Step-by-step explanation:

REMEMBER: Area is found by multiplying one side, by the other side that isn't the same length as it.

Multiply 12 by 10 to get 120

120 is the area of the bedroom

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2 years ago
I dont know how to do this.
Aleksandr-060686 [28]
The answer to the question

5 0
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What are the two smallest consecutive odd integers whose sum is greater than 17
Mice21 [21]

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9 and 11

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2 years ago
<img src="https://tex.z-dn.net/?f=%5Clongmapsto%5Csqrt%7B64%7D" id="TexFormula1" title="\longmapsto\sqrt{64}" alt="\longmapsto\s
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\sqrt{64}  = 2 \times 2 \times 2 \times 2 \times 2 \times 2

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<h3># Answered by Cutest Ghost..</h3>

7 0
2 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

7 0
3 years ago
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