Given:
Sum of two consecutive integers is 325.
To find:
The number that is the least of the two consecutive integers.
Solution:
Let the two consecutive integers are x and x+1. So,
![x+(x+1)=325](https://tex.z-dn.net/?f=x%2B%28x%2B1%29%3D325)
![2x+1=325](https://tex.z-dn.net/?f=2x%2B1%3D325)
Subtract 1 from both sides.
![2x=324](https://tex.z-dn.net/?f=2x%3D324)
Divide both sides by 2.
![x=\dfrac{324}{2}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B324%7D%7B2%7D)
![x=162](https://tex.z-dn.net/?f=x%3D162)
The second integer is
![x+1=162+1=163](https://tex.z-dn.net/?f=x%2B1%3D162%2B1%3D163)
Since
, therefore, the least of the two consecutive integers is 162.
Square root of 21 is
<span>4.58257569496
</span>
I rounded it to 4.58
A^2 + a - 6 here's your answer mate, hope it helps
I'm really sorry, but I only know the first question.
The answer is 340.
Answer:
2/3pix³ or ![\frac{2}{3} \pi x^3\\](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D%20%5Cpi%20x%5E3%5C%5C)
Step-by-step explanation:
This problem brothers on the mensuration of solid shapes, a cone.
we know that the expression for the volume of a cone is
![volume= \frac{1}{3} \pi r^2h](https://tex.z-dn.net/?f=volume%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20r%5E2h)
let the radius r of the cone be= ![x](https://tex.z-dn.net/?f=x)
and the height h is =![2x](https://tex.z-dn.net/?f=2x)
we can now solve the expression that represents the volume of the cone, in cubic units.
![volume= \frac{1}{3} \pi *x*2x\\\volume= \frac{1}{3} \pi *2x^3\\\volume= \frac{2}{3} \pi x^3\\](https://tex.z-dn.net/?f=volume%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20%2Ax%2A2x%5C%5C%5Cvolume%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20%2A2x%5E3%5C%5C%5Cvolume%3D%20%5Cfrac%7B2%7D%7B3%7D%20%5Cpi%20x%5E3%5C%5C)