proyaop

If the rectangular frame has width x, and length y, then the perimeter is 2x + 2y = 39. The length (y) is 2.5 more than the width (x), so y = x + 2.5

Then you can substitute it back into the first equation. 2x + 2(x + 2.5) = 39. 4x + 5 = 39. 4x = 34, x = 17/2. y = 8.5 + 2.5 = 11

The width is 8.5 inches, and the height is 11 inches.

The equation can be written to standard form. 

x/3 + y/4 = 6        =>     4x + 3y = 72         => point-slope form =>     y = -4x/3 + 24

Then since our triangle has base of y = 0 and height of x = 0, we can do some calculations to see the area under the line y = -4x/3 + 24 in Quad 1.

It has y-intercept of 24. x - intercept is 0 = -4x/3 + 24. 0 = -4x + 72. x = 18. So the x-intercept is (18, 0)

Thus the base has length 18, and the height has length 24, the area is 18 * 24 / 2 = 216.

Both sides of the equation have a common denominator, so you can multiply both sides by (a^3 + 7) to get rid of it.

You are left with 1 = -(a^3). Since a is a real number, then the only solution is a = -1.

You can do p + 2p = 3p + 3

Just try and get ap + bp = (a + b)p + c where c is not = 0.

If its perpendicular, then the slopes of the two lines multiply to negative 1.

y = 3x/2 + b is what we know so far.

Plug in 4, -3

-3 = 6 + b

b = -9

So the equation is y = 3x/2 - 9

Set line PQ to be the x-axis (just for visualization, not going to use coordinate bashing).

The point of tangency on circle P is F, and the point of tangency on circle Q is S.

PQ has length 18. PF is the radius, so it has length 10, and QS is the radius, so it has length 8.

Let the point of intersection between line PQ and line FS be O.

Let QO = x (looking for x).

PF is perpendicular to FS is perpendicular to QS. So we have similar triangles OQS and OPF.

Similar triangles: x/(x + 18) = 8/10

10x = 8x + 144

QR = 72

First combine like terms: x^2 + y^2 + 12x - 10y - 38 = 0

Complete the square: x^2 + 12x + 36 - 36 + y^2 - 10y + 25 - 25 - 38 = 0   (doing this to obtain square of binomials, remember equation for circle)

(x + 6)^2 + (y - 5)^2 = 99 = r^2

Area of a circle = pi * r^2             =>        99pi

3*3*y = 3*3*2*27*4*81*2

cancel out 3*3

y = 2*27*4*81*2 = 34992

Cancel stuff out on each side.

5*13*_*3 = 17   => 195 * _ = 17

_ = 17/195

3y = 7x + 2

Set x = 0

y = 2/3

(0, 2/3)