Well, I suppose you could break up the coordinate plane
in millions of different ways.
If you take each region to be all the space that is not crossed
by an axis, then you have the four Quadrants.
25t means "25 times t" where t is some unknown number. It is a placeholder for a number.
To find what the number is, we undo what is happening to t. So we divide both sides by 25 to undo the operation "multiply by 25"
----------
25*t = 1125
25*t/25 = 1125/25 divide both sides by 25
t = 45
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<h3>Answer: 45</h3>
As a check, we can plug t = 45 into the equation and we should get the same value on both sides
25*t = 1125
25*45 = 1125 replace every t with 45
1125 = 1125 the answer is confirmed
Answer:
in
Step-by-step explanation:
Let x be the side of square.
Length of box=8-2x
Width of box=15-2x
Height of box=x
Volume of box=![L\times B\times H](https://tex.z-dn.net/?f=L%5Ctimes%20B%5Ctimes%20H)
Substitute the values then we get
Volume of box=V(x)=![(8-2x)(15-2x)x=(15x-2x^2)(8-2x)](https://tex.z-dn.net/?f=%288-2x%29%2815-2x%29x%3D%2815x-2x%5E2%29%288-2x%29)
![V(x)=120x-30x^2-16x^2+4x^3](https://tex.z-dn.net/?f=V%28x%29%3D120x-30x%5E2-16x%5E2%2B4x%5E3)
![V(x)=4x^3-46x^2+120x](https://tex.z-dn.net/?f=V%28x%29%3D4x%5E3-46x%5E2%2B120x)
Differentiate w.r.t x
![V'(x)=12x^2-92x+120](https://tex.z-dn.net/?f=V%27%28x%29%3D12x%5E2-92x%2B120)
![V'(x)=0](https://tex.z-dn.net/?f=V%27%28x%29%3D0)
![12x^2-92x+120=0](https://tex.z-dn.net/?f=12x%5E2-92x%2B120%3D0)
![3x^2-23x+30=0](https://tex.z-dn.net/?f=3x%5E2-23x%2B30%3D0)
![3x^2-18x-5x+30=0](https://tex.z-dn.net/?f=3x%5E2-18x-5x%2B30%3D0)
![3x(x-6)-5(x-6)=0](https://tex.z-dn.net/?f=3x%28x-6%29-5%28x-6%29%3D0)
![(x-6)(3x-5)=0](https://tex.z-dn.net/?f=%28x-6%29%283x-5%29%3D0)
![x-6=0\implies x=6](https://tex.z-dn.net/?f=x-6%3D0%5Cimplies%20x%3D6)
![3x-5=0\implies x=\frac{5}{3}](https://tex.z-dn.net/?f=3x-5%3D0%5Cimplies%20x%3D%5Cfrac%7B5%7D%7B3%7D)
Again differentiate w.r.t x
![V''(x)=24x-92](https://tex.z-dn.net/?f=V%27%27%28x%29%3D24x-92)
Substitute x=6
![V''(6)=24(6)-92=52>0](https://tex.z-dn.net/?f=V%27%27%286%29%3D24%286%29-92%3D52%3E0)
Substitute x=5/3
![V''(5/3)=24(5/3)-92=-52](https://tex.z-dn.net/?f=V%27%27%285%2F3%29%3D24%285%2F3%29-92%3D-52%3C0)
Hence, the volume is maximum at x=![\frac{5}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B3%7D)
Therefore, the side of the square ,
in cutout that gives the box the largest possible volume.
The square root of 69 is 8.30.because 8.30*8.30 is 69