Let the least possible value of the smallest of 99 cosecutive integers be x and let the number whose cube is the sum be p, then
![\frac{99}{2} (2x+98)=p^3 \\ \\ 99x+4,851=p^3\\ \\ \Rightarrow x=\frac{p^3-4,851}{99}](https://tex.z-dn.net/?f=%20%5Cfrac%7B99%7D%7B2%7D%20%282x%2B98%29%3Dp%5E3%20%5C%5C%20%20%5C%5C%2099x%2B4%2C851%3Dp%5E3%5C%5C%20%5C%5C%20%5CRightarrow%20x%3D%5Cfrac%7Bp%5E3-4%2C851%7D%7B99%7D)
By substitution, we have that
![p=33](https://tex.z-dn.net/?f=p%3D33)
and
![x=314](https://tex.z-dn.net/?f=x%3D314)
.
Therefore, <span>the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube is 314.</span>
Mark and Don have to sell 11 and 59 marbles respectively.
<em><u>Explanation</u></em>
Lets assume, the number of marbles Mark has is ![x](https://tex.z-dn.net/?f=x)
As, Don has 4 more than 5 times the number of marbles mark has, so the number of marbles Don has ![= 5x+ 4](https://tex.z-dn.net/?f=%3D%205x%2B%204)
Question says that the total number of marbles is 70. So, the equation will be...
![x+(5x+4)= 70\\ \\ 6x+4= 70 \\ \\ 6x = 70-4\\ \\ 6x=66 \\ \\ x= \frac{66}{6}= 11](https://tex.z-dn.net/?f=x%2B%285x%2B4%29%3D%2070%5C%5C%20%5C%5C%206x%2B4%3D%2070%20%5C%5C%20%5C%5C%206x%20%3D%2070-4%5C%5C%20%5C%5C%206x%3D66%20%5C%5C%20%5C%5C%20x%3D%20%5Cfrac%7B66%7D%7B6%7D%3D%2011)
<em>So, the number of marbles Mark has = 11 </em>
<em>and the number of marbles Don has =
</em>
Answer: It will be C, 5x + 2y = 6 is equivalent to y = -5/2x + 3
Answer:
6 StartRoot 3 EndRoot units
Step-by-step explanation: