jfan17

https://web2.0calc.com/questions/reposted-again-bc-a-certain-user-kept-complaining#r7

I believe I posted an answer to your first question already. You can find the solution here:

https://web2.0calc.com/questions/reposted-again-bc-a-certain-user-kept-complaining#r7

let me know if you see anything wrong with my solution, and I'll do my best to correct it.

Also, don't repost your question after it hasn't been answered for an hour or two. That's unnecessary. There just may not be people on at this time.

As it was written by the Guest, ako correctly points out that this equation has no solutions. However, I will take it that they forgot parentheses(important!) around the denominator and numerator. I'll then solve it with this interpretation:

\({x\over x-5} = {4 \over x-4}\)

Cross multiplying ,we get

\(4x-20 = x^2-4x\)

Rearranging, we get:

\(x^2-8x + 20 = 0\)

Realizing that our roots are imaginary because of the discriminant(but the problem never limits the value of x, so I'll continue), we can use the quadratic formula to find:

\(x = {8 \pm \sqrt{-16} \over 2}\)

\(x = {8 \pm 4i \over 2} = 4\pm2i\)

Because we have the roots of this quadratic given as : 

\(x = {-5 \pm \sqrt{c} \over 2}\), by vietas, we can write:

\({-5 + \sqrt{c} \over 2} * {-5 - \sqrt{c} \over 2}\) = c

Expanding, we get:

\({25-c\over4} = c\), because the square roots cancle out

then, this gives us:

\(4c = 25-c\) when we multiply by 4 on both sides

\(5c = 25\)

Dividing by 5 on both sides, we get:

\(c = 5\)

By vietas formulas:

a + b = -(-1)/1 = 1/1 = 1

Vietas formulas state:

Given a quadratic(it can be any polynomial, I'd recommend searching up vieta's formulas)

\(ax^2+bx+c = 0\)

and roots: x₁, x₂

x₁ + x₂ = \(\frac{-b}a\)

x₁*x₂ = \(\frac ca\)

The google sites doesn't work for me..... It says:

404. That’s an error.

The requested URL was not found on this server. That’s all

Use the fact that:

sin(x) = cos(90-x)

We then can rewrite our equation as:

cos²(0) + cos²(1) + cos²(2)........ cos²(44) + sin²(45) + sin²(44) ........ sin²(0).

Next we make use of a basic trig identity:

\(\cos^2\theta + \sin^2\theta = 1\)

Given that the two angles are the same.

We realize we have 0-44 = 45 total such pairs in our rewritten equation(can you see how?).

This gives us 45 + sin²(45) =  \(45+(\sqrt2/2)^2 = 45 + 2/4 = 45 + 1/2 = 45.5\)

What's the question?

Slope is equivalent to rise over run, or the change in y / change in x between two given (distinct) points on the line\(\). Let's look at the x and y intercepts which are 10 and -6 respectively. Over a distance of 10 units, the graph rises by 6, meaning that rise / run = 6/10 = a slope of 3/5

Hey guest! let's first write out what we're given:

\(\frac{a-b}a = \frac b{a-b}\)

Cross multiplying, we can write:

\(ab = (a-b)^2\)

Expand this out, and we get:

\(ab = a^2-2ab + b^2\)

Subtracting ab on both sides, we then get:

\(a^2-3ab + b^2 = 0\)

Now, we divide by b²(and you'll see why pretty soon). This gives us:

\((\frac ab)^2 -3(\frac a b) + 1 = 0\)

Let's then name \(\frac a b\)as x(Do you see why? Because it makes for easy substitution for a quadratic. If we define it as x, if we solve for the values of x, we solve for the values of \(\frac a b\))

Substituting into our quadratic, we get:

\(x^2-3x+1 = 0\)

Clearly, this isn't factorable with integer roots, so we use the quadratic formula. This then gives us:

\((-(-3)\pm\sqrt{(-3)^2-4*1*1})/2 = (3\pm\sqrt{9-4})/2 = (3\pm\sqrt{5})/2\) as our two possible values of a/b. However, we're not done yet. The question asks us for the sum of all possibles values for \(\frac ab\). We can see that our two roots are the conjugates of each other, meaning that their irrational part cancels out. This then leaves us with:

\(\frac {3+ \sqrt5} {2} + \frac{3-\sqrt5} {2} = \frac 6 2 = 3\) as our final answer