Answer:
13.34 pm
22.43 pm
16.22 pm
i cant see the last one clearly
Answer:20 girls
Step-by-step explanation:the difference between 3 and 4 is one so there will be a difference of one between the 2 questions.
The sum of 5 and the quotient of 4 and 2 would be the answer
To find 
, you can switch the "x" and "f(x) or y" in the equation.
![f(x) = \sqrt[3]{x-2} +8](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx-2%7D%20%2B8)
![y = \sqrt[3]{x-2}+ 8](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7Bx-2%7D%2B%208)
![x = \sqrt[3]{y-2}+8](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7By-2%7D%2B8)
Now you need to isolate the "y"
Subtract 8 on both sides
![x - 8 = \sqrt[3]{y-2}](https://tex.z-dn.net/?f=x%20-%208%20%3D%20%5Csqrt%5B3%5D%7By-2%7D)
Cube ( ³ ) each side to get rid of the ∛
![(x-8)^{3} = (\sqrt[3]{y-2}) ^{3}](https://tex.z-dn.net/?f=%28x-8%29%5E%7B3%7D%20%3D%20%28%5Csqrt%5B3%5D%7By-2%7D%29%20%5E%7B3%7D)
Add 2 on both sides
<h3>
Answer: Always</h3>
Proof:
Let
- A = some rational number
- B = some irrational number
- C = some rational number (that isn't necessarily equal to A)
We'll show that a contradiction happens if we tried to say A+B = C.
In other words, we'll show a contradiction happens for the form rational+irrational = rational.
Because A and C are rational, we can say
A = p/q
C = r/s
where p,q,r,s are integers. The q and s in the denominators cannot be zero.
So,
A+B = C
B = C - A
B = (r/s) - (p/q)
B = (qr/qs) - (ps/qs)
B = (qr-ps)/(qs)
B = (some integer)/(some other integer)
B = some rational number
But wait, we stated that B was irrational. The term "irrational" literally means "not rational". Irrational numbers cannot be written into fraction form of two integers. This is a contradiction and shows that A+B = C is never possible if A,C are rational while B is irrational.
This must lead to the conclusion that A+B must always be irrational if A is rational and B is irrational.
The template is this
rational + irrational = irrational
rational + rational = rational
