A binomial squared is written like
So, we have 
So, we want to complete the square 
We can add and subtract 9 to write
V of w sphere =
r³
Multiply by 3 on either sides to get rid of the fraction.
3V = 4
r³
Now divide either sides by 4
to isolate r³
r³
4
and 4
cancels out
= r³
Take the cube root to isolate r.
![\sqrt[3]{ \frac{3V}{4 \pi } } = \sqrt[3]{r^3}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3V%7D%7B4%20%5Cpi%20%7D%20%7D%20%3D%20%20%5Csqrt%5B3%5D%7Br%5E3%7D%20)
the cube root cancels the cube
![\sqrt[3]{ \frac{3V}{4 \pi } }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3V%7D%7B4%20%5Cpi%20%7D%20%7D%20)
= r
Answer:
I am pretty sure its y={5}/{4}x-5
Step-by-step explanation:
As x approaches 5
hmm, we can divide the (x-5) from top and bottom to get x-5
if we input 5 for x we get
5-5=0
it approaches 0 as x approaches 5
Answer:
You are given:
4Fe+3O_2 -> 2Fe_2O_3
4:Fe:4
6:O_2:6
You actually have the same number of Fe on both sides, The same is true for O_2 so yes this equation is properly balanced.
For added benefit consider the following equation:
CH_4+O_2-> CO_2+2H_2O
ASK: Is this equation balanced? Quick answer: No
ASK: So how do we know and how do we then balance it?
DO: Count the number of each atom type you have on each side of the equation:
1:C:1
4:H:4
2:O:4
As you can see everything is balanced except for O To balance O we can simply add a coefficient of 2 in front of O_2 on the left side which would result in 4 O atoms:
CH_4+color(red)(2)O_2-> CO_2+2H_2O
1:C:1
4:H:4
4:O:4
Everything is now balanced.
Step-by-step explanation:
