Answer:
![\boxed{ \bold{ \huge{ \boxed{ \sf 240 \: {inches}^{2} }}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%20%5Chuge%7B%20%5Cboxed%7B%20%20%5Csf%20240%20%5C%3A%20%20%7Binches%7D%5E%7B2%7D%20%7D%7D%7D%7D)
Step-by-step explanation:
Given,
Length of a rectangle = 20 inches
Perimeter of a rectangle = 64 inches
Area of a rectangle = ?
Let width of a rectangle be ' w ' .
<u>Fi</u><u>rst</u><u>,</u><u> </u><u>finding </u><u>the</u><u> </u><u>width</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>rectangle</u>
![\boxed{ \sf{perimeter = 2(l + w)}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%7Bperimeter%20%3D%202%28l%20%2B%20w%29%7D%7D)
plug the values
⇒![\sf{64 = 2(20 + w)}](https://tex.z-dn.net/?f=%20%5Csf%7B64%20%3D%202%2820%20%2B%20w%29%7D)
Distribute 2 through the parentheses
⇒![\sf{64 = 40 + 2w}](https://tex.z-dn.net/?f=%20%5Csf%7B64%20%3D%2040%20%2B%202w%7D)
Swap the sides of the equation
⇒![\sf{40 + 2w = 64}](https://tex.z-dn.net/?f=%20%5Csf%7B40%20%2B%202w%20%3D%2064%7D)
Move 2w to right hand side and change it's sign
⇒![\sf{2w = 64 - 40}](https://tex.z-dn.net/?f=%20%5Csf%7B2w%20%3D%2064%20-%2040%7D)
Subtract 40 from 64
⇒![\sf{2w = 24}](https://tex.z-dn.net/?f=%20%5Csf%7B2w%20%3D%2024%7D)
Divide both sides of the equation by 2
⇒![\sf{ \frac{2w}{2} = \frac{24}{2} }](https://tex.z-dn.net/?f=%20%5Csf%7B%20%5Cfrac%7B2w%7D%7B2%7D%20%20%3D%20%20%5Cfrac%7B24%7D%7B2%7D%20%7D)
Calculate
⇒![\sf{w = 12 \: inches}](https://tex.z-dn.net/?f=%20%5Csf%7Bw%20%3D%2012%20%5C%3A%20inches%7D)
Width of a rectangle ( w ) = 12 inches
<u>Now</u><u>,</u><u> </u><u>finding</u><u> </u><u>the</u><u> </u><u>area</u><u> </u><u>of </u><u>a</u><u> </u><u>rectangle</u><u> </u><u>having</u><u> </u><u>length</u><u> </u><u>of</u><u> </u><u>2</u><u>0</u><u> </u><u>inches</u><u> </u><u>and </u><u>width </u><u>of</u><u> </u><u>1</u><u>2</u><u> </u><u>inches</u>
![\boxed{ \sf{area \: of \: rectangle = length \: \times \: \: width}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%7Barea%20%5C%3A%20of%20%5C%3A%20rectangle%20%3D%20length%20%5C%3A%20%20%5Ctimes%20%20%5C%3A%20%5C%3A%20width%7D%7D)
plug the values
⇒![\sf{area \: of \: rectangle =20 \times 12 }](https://tex.z-dn.net/?f=%20%5Csf%7Barea%20%5C%3A%20of%20%5C%3A%20rectangle%20%3D20%20%5Ctimes%20%2012%20%7D)
Multiply the numbers : 20 and 12
⇒![\sf{area \: of \: rectangle = 240 \: {inches}^{2} }](https://tex.z-dn.net/?f=%20%5Csf%7Barea%20%5C%3A%20of%20%5C%3A%20rectangle%20%3D%20240%20%5C%3A%20%20%7Binches%7D%5E%7B2%7D%20%7D)
Hence, Area of a rectangle = 240 inches²
Hope I helped !
Best regards!
Answer:
75.44 Square Inches
Step-by-step explanation:
The diagram of the problem is produced and attached.
To determine the area of the cleaned sector:
Let the radius of the larger sector be R
Let the radius of the smaller sector be r
Area of the larger sector ![=\frac{\theta}{360}X\pi R^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctheta%7D%7B360%7DX%5Cpi%20R%5E2)
Area of the smaller sector ![=\frac{\theta}{360}X\pi r^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctheta%7D%7B360%7DX%5Cpi%20r%5E2)
Area of shaded part =Area of the larger sector-Area of the smaller sector
![=\frac{\theta}{360}X\pi R^2-\frac{\theta}{360}X\pi r^2\\=\frac{\theta \pi}{360}X (R^2- r^2)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctheta%7D%7B360%7DX%5Cpi%20R%5E2-%5Cfrac%7B%5Ctheta%7D%7B360%7DX%5Cpi%20r%5E2%5C%5C%3D%5Cfrac%7B%5Ctheta%20%5Cpi%7D%7B360%7DX%20%28R%5E2-%20r%5E2%29)
From the diagram, R=10 Inch, r=10-7=3 Inch, ![\theta=95^\circ](https://tex.z-dn.net/?f=%5Ctheta%3D95%5E%5Ccirc)
Therefore, Area of the sector cleaned
![=\frac{95 \pi}{360}X (10^2- 3^2)\\=75.44$ Square Inches](https://tex.z-dn.net/?f=%3D%5Cfrac%7B95%20%5Cpi%7D%7B360%7DX%20%2810%5E2-%203%5E2%29%5C%5C%3D75.44%24%20Square%20Inches)
Ratio of cat : to dog = 40 : 20 = 2:1
Answer:
39 + 5d
Step-by-step explanation:
First multiply 5 by 9 and 5 by d then subtract 6 from 45 (which you get from multiplying 5 by 9) and you get 39 and multiply 5 by d so you get 39 + 5d
Answer:
m = -10
Step-by-step explanation:
m = y2 - y1 / x2 - x1
m = 60-50 / -2 - -1
m = 10/ -1
m = -10