Answer:
You need to have a labotomy sir
Step-by-step explanation:
Let , smallest integer is x.
So , other one is x + 1.
By , given conditions :
![x( x + 1 ) = 8x + 18\\\\x^2+x=8x+18\\\\x^2-7x-18=0\\\\x^2-9x+2x-18=0\\\\x(x-9)+2(x-9)=0\\\\x=-2 \ ,\ x = 9](https://tex.z-dn.net/?f=x%28%20x%20%2B%201%20%29%20%3D%208x%20%2B%2018%5C%5C%5C%5Cx%5E2%2Bx%3D8x%2B18%5C%5C%5C%5Cx%5E2-7x-18%3D0%5C%5C%5C%5Cx%5E2-9x%2B2x-18%3D0%5C%5C%5C%5Cx%28x-9%29%2B2%28x-9%29%3D0%5C%5C%5C%5Cx%3D-2%20%5C%20%20%2C%5C%20x%20%3D%209)
Since, the numbers are positive so, x = -2 is ignored.
Therefore, the numbers are 9 and 10.
Hence, this is the required solution.
Answer:
I believe everything is correct
Step-by-step explanation:
Hope That Helped :)
A
Since you can carry 0 boxes but max is 10 and you cant carry 2/3 of a box or a square root of a box, they have to be whole boxes (whole integers)
Ummm... I explained the whole process earlier, and was mistakenly issued for plagiarism. Which really sucks, because it took forever to type and format all of those formulas and steps.
Here's the shortened version...
V=
![4/3 \pi r^{3}](https://tex.z-dn.net/?f=4%2F3%20%5Cpi%20r%5E%7B3%7D%20)
d (diameter) = 2r
so 24/2 = 12
plug in...
You should get approx. 7234.56, and don't forget to label it appropriately, it should be units cubed.