proyaop

Add the values of the coordinates of the two points, then divide it by 2. Thus, the midpoint of the segment would be ((-1 - 6)/2, (5 - 2)/2)

Thus, the coordinates of the midpoint of the line segment is (-3.5, 1.5).

:)

Triangle CHB is similiar to triangle GFB by angle angle similarity (shared angle B and right angle).

Thus, the ratio of CH to GF is 18:3 or 6:1. Since CH is 24, then GF is 4. 

If GF is 4, then the area of triangle GFB is 6. The area of triangle ADE is also 6. To calculate the area of the pentagon, just calculate the area of the entire trianglce ABC and subtract it by ([ADE] + [GFB]). The area of triangle ABC is 36 x 24 / 2 = 432 units^2.

Thus, the area of pentagon CDEFG is 420 units^2.

If a chain weighs 10 pounds per foot, and a foot is 12 inches, and 16 ounces is a pound, then 4 inches is 10/3 pounds. 

Since 16 ounces are in a pound, there are 16 x 10/3 ounces in 4 inches of chain.

4 inches of chain weigh 160/3 ounces or 53.3333 ounces.

Since triangle ADC is an isosceles triangle, then angle DAC = angle ACD = 24 degrees. Thus, angle ADB = angle ACD + angle DAC by exterior angle theorem, so angle ADB = 24 + 24 = 48 degrees. Since triangle ABD is an isosceles triangle, then angle ABD = angle ADB = 48 degrees.

Angle ABC = angle ABD.

Angle ABC = 48 degrees.

I am assuming that we have a triangle ABC even though not posted in the question. If that is so, then the first step is to identify a few extra triangles we can draw in. 

I drew an altitude from point C down to AB, and called the point of intersection D. 

Angle A is 45 degrees, so triangle ADC is a right isosceles triangle with hypotenuse AC. 

Angle B is 30 degrees, so triangle BCD is a 30-60-90 triangle with hypotenuse BC.

If those are the cases, then DC is \({\sqrt{6}\over2}\). We also know that AD = DC, so AD = \({\sqrt{6}\over2}\). BD is \(\sqrt{3}\) x DC, so BD is \(3\sqrt{2}\over2\).

What we also do know is that AB = BD + AD = \({\sqrt{6}\over2} + {3\sqrt{2}\over2}\).

Thus, \(AB = {\sqrt{6} + 3\sqrt{2}\over2}\).

PRYAP

First I simplified the first equation to:

x + y = 5xy. I got this by multiplying both sides of the equation by 'xy'.

Then we can substitute that into the second equation, getting:

3xy + 5xy = 4 + 5xy.

This simplifies to 3xy = 4.

And xy = 4/3.

Since the question is looking for x^2y + xy^2, that is equal to: (x+y)(xy).

Substituting the value back in, we have (5xy)(xy), and finally:

\(x^2y + xy^2 = {80\over9}\)

This analytical geometry question has been answered correctly here: https://web2.0calc.com/questions/analytic-graphing-geometry-help

Also, the perimeter wouldn't change whether it would be cut by a different line because they are two CONGRUENT quadrilaterals, and maybe there was a typo for y = x/3, not y = x/2. Either way, go check out the URL :D

good luck with your alcumus.

To make the equation clear, it looks like this: \({y\over3}+1={y+3\over y}+7\).

Then multiply every term by '3y'.

It should now look like this:\(y^2 + 3y = 3y + 9 + 21y\).

Combining like terms and moving the equation to one side, we get: \(y^2 - 21y - 9 = 0\).

Then you can use the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) where a is the coefficient of x^2. b is the coefficient of x, and c is the constant.

Plugging in the values, we have:

\(y = {21 + 3\sqrt{53}\over2}\)

\(y = {21-3\sqrt{53}\over2}\)

Those are the two possible values of y :)

So first we make the coefficient of x^2 = 1.

The new quadratic is: \(x^2 + {5\over3}x + {7\over3} = 0\)

Then since we know by Vieta, the two solutions of the quadratic multiply to the c term, and they add to the b term * -1. The solutions are u and v.

Thus, \(u + v = -{5\over3}\), and \(uv = {7\over3}\). 

Since \(u^2 + v^2 = (u + v)^2 - 2uv\), we can substitute in our previous values. 

Then \(u^2 + v^2 = {25\over9} - {14\over3}\).

Thus, \(u^2 + v^2 = -{17\over9}\).

Wow you are all right! Great strategies for encountering this hard problem...

Thanks!