Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.
think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.
now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area. Well, what is the circumference of that shaded area?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D12%20%5Cend%7Bcases%7D%5Cimplies%20C%3D2%5Cpi%2012%5Cimplies%20C%3D24%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bthree%20quarters%20of%20it%7D%7D%7B24%5Cpi%20%5Ccdot%20%5Ccfrac%7B3%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%206%5Cpi%20%5Ccdot%203%5Cimplies%2018%5Cpi)
well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20C%3D18%5Cpi%20%5Cend%7Bcases%7D%5Cimplies%2018%5Cpi%20%3D2%5Cpi%20r%5Cimplies%20%5Ccfrac%7B18%5Cpi%20%7D%7B2%5Cpi%20%7D%3Dr%5Cimplies%209%3Dr%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bdiameter%20is%20twice%20the%20radius%7D%7D%7Bd%3D18%7D~%5Chfill)
<span><span>1.
</span>Lucas Lives 5/8 miles from their schools
=> 5/8 miles is equals to:
=> 5 / 8 = 0.625 miles.
Now, the problem says that Kenny lives twice as far as Lucas from their school.
Therefore, we only need to multiply the total number of miles of Lucas distance
from house to school to be able to get what is the distance of Kenny’s house to
school
=> 0.625 * 2 = 1.25 miles or 1 and ¼ miles.</span>
35z^4 + 11z^4<span>
= 46z^4 ...combine like terms
hope that helps</span>
Answer:
72
Step-by-step explanation:
The area (A) of a rhombus is calculated as
A =
× d₁ × d₂ (d₁ and d₂ are the diagonals )
The diagonals bisect each other at right angles
d₁ = 2 × 6 = 12
Use the tangent ration in the upper left right triangle and the exact value
tan60° = 
tan60° =
=
=
( multiply both sides by 6 )
opp = 6
, then
d₂ = 2 × 6
= 12
Thus
A =
× 12 × 12
= 6 × 12
= 72