Given:
A rectangle has a length of 8 and a width of 4.
Four equal rectangles are drawn within that rectangle, creating the longest lengths possible for the rectangles.
To find:
The sum of the perimeters of two of these equal rectangles.
Solution:
We have to draw 2 lines which are bisecting the length and width respectively, to divide a rectangle in 4 equal parts with longest lengths.
Half of length 8 = 4
Half of width 4 = 2
It means each rectangle have a length of 4 and a width of 2.
Perimeter of a rectangle is
![Perimeter=2(length+width)](https://tex.z-dn.net/?f=Perimeter%3D2%28length%2Bwidth%29)
![Perimeter=2(4+2)](https://tex.z-dn.net/?f=Perimeter%3D2%284%2B2%29)
![Perimeter=2(6)](https://tex.z-dn.net/?f=Perimeter%3D2%286%29)
![Perimeter=12](https://tex.z-dn.net/?f=Perimeter%3D12)
So, perimeter of each equal rectangle is 12 units.
Sum of the perimeters of two of these equal rectangles is
![12+12=24](https://tex.z-dn.net/?f=12%2B12%3D24)
Therefore, sum of the perimeters of two of these equal rectangles is 24.
We need to perform division of polynomials such as the solution is shown below:
( 32x8 - 8x6 + 28x4) / 4x4
(32x8 / 4x4) - ( 8x6 / 4x4 ) + (28x4 / 4x4)
8x4 - 2x2 + 7
The power of the result was already arranged in descending order.
The answer is "8x4 - 2x2 + 7 ".
Answer:
○![\displaystyle \frac{120}{169}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B120%7D%7B169%7D)
Explanation:
Looking in quadrant I, in the degree-angle range from 0 to
take the <em>inverse of </em><em>sine</em> to get your angle measure, then plug it into the other <em>sine</em> function to get the above answer.
I am joyous to assist you anytime.
Answer:
y=.75 x=.75 for one cup of water
Step-by-step explanation: