aboslutelydestroying

I don't think the answer is right

Do you think you might of did something wrong?

Pluging in random numbers for n works, but can anyone find a better way to get the possible values of n? Thank you

https://web2.0calc.com/questions/in-triangle-pqr-m-is-the-midpoint-of-pq-let-x-be_1

Thannks so much!

To find the constant term in the expansion of the expression (2z - 1/√z) * 5 *(z - 1/√z)^3, we first need to expand the entire expression using the binomial theorem. Start by simplifying this equation to get a polynomial. However, we’re only interested in the constant term, so we’re looking for terms that won’t include z when multiplied together. We find these in the following multiplicands from each factor: -1/√z from (2z - 1/√z), and z^3 from (z - 1/√z)^3. When these are multiplied together, they result in a constant term.

So the answer is: \(-1/\sqrt{z} \) and \(z^3\)

\(x^{14}+4x^{11}+6x^8+4x^5+x^2\)

Is the answer

Thanks so much CPhill, you are a really good answerer

For number 4)

\(x+y+3z=10\)

\(4x+2y+5z=7\)

\(kx+z=3\)

Multiply 2 on both sides of the 1st equation:

\(2x+2y+6z=20\)

Subtract it to the 2nd equation: 

\(2x-z=-13\)

\(kx+z=3\)

multiply -1 to \(kx+z=3\) on both sides:

\(-kx-z=-3\)

in order for this system to have no solutions

k has to be -2 which gets us: \(2x-z=-3\)

This equation can't have 2 values since it is linear, 

so this proof tells us, k = 2

For number 3)

\(\sqrt{4x^2}+\sqrt{x^2}=6\)

\(|2x|+|x|=6\)

Say 2x+x = 6

3x = 6

x = 2

Then there is the other case:

2x+x = -6

3x = -6

x = -2

So the values of x would be -2 and 2