gb1falcon

The fractional form of 0.8 is \(\frac{4}{5}\) because \(\frac{4}{5}=0.8\).

If x2 is \(x^2\) then use the quadratic formula.

 \(x^2+9x+16=0\)

\(x = {-9 \pm \sqrt{9^2-4 \cdot 1 \cdot 16} \over 2 \cdot 1}\)

\(x = {-9 \pm \sqrt{81-64} \over 2}\)

\(x = {-9 \pm \sqrt{17} \over 2}\)

Since this is irrational, the expression cannot be factored.

The three diagonals define APB to be a 45-45-90 triangle, so x=\(\sqrt{2}\).

I'm glad you could give me this answer. Thanks!

Could someone show me a detailed way to solve this problem so I can use it elsewhere?

Add equations together to get 15(a+b+c+d+e)=215.

Divide to get a+b+c+d+e=43/3.

Now take equation 1, subtract it from equation 2.

a+b+c+d-4e=-26. Subtract the two equations shown to get 5e=121/3, so e=121/15.

See if you can use this same method to determine a, b, c, and d.

I like that approach, but I don't think it is entirely correct. In this problem, \(|z-1|=|x+yi-1|=|(x-1)+yi|=\sqrt{(x-1)^2+y^2}\), because \(|x+yi|\) signifies the magnitude of the number, which is \(\sqrt{x^2+y^2}.\) I think that you interpreted it as absolute value.

Good rounding. I'm think that the expression winds up being 2*1.5=3.

I don't know why, but that is what I get.

\(\frac{3x + 12}{x + 4} = (4x^2 + 8x + 18x^2)(2x)\)

\(3=(4x^2 + 8x + 18x^2)(2x)\)

\(3=(2x)(22x^2+8x)\)

\(3=44x^3+16x^2\)

I'm afraid that's as far as I can get you.

There is no such number. I went through and tried all of the possible numbers, none of them work.

Here is a short way to do so: If the digits are within 4 of each other, then the number's reversal will be too close. However, if they are 6 or above away from each other, the number's reversal will be too far away. So the digits must be 5 apart. Since the first digit must be larger, the possibilities are 50,61,72,83, and 94. None of these match the last requirement, so there is no such number.