tertre

If you see closely, the line intersects at (0,0)-making this both the x and y-intercepts.

1. For this problem, recognize that the given side lengths are part of an isoceles triangle. 

It is advisable to drop the altitude to the base length of six, bisecting that side length.

By the Pythagorean Theorem, we have 7^2-3^2=y^2, y^2=40, \(y=2\sqrt{10}\).

Thus, the area of the triangle is \(\frac{1}{2}*6*2\sqrt{10}=6\sqrt{10}\)

82 should be the denominator.

I tried rewriting the bases as powers of three.

Then, I tried diving by three and multiplying the series by three

but it did not work.

27/27 looks very fishy, might be the key to the problem.

I have a hunch, but it probably makes no sense.

Maybe the use of Mass Points works here.

Set a mass of two of point Q, one for M and one for P.

If B has a mass of one, then point A must have a different mass, and the two masses add to some number.

However, both mass A and mass C add to point P, which has a mass of one.

D has a mass greater than one.

Finally, that means the ACD triangle area is greater than that of ABP triangle.

I'm sure that there is a way better explanation.

According to the information given in the problem....we have a^2*b^3=y, y is a random constant.

Thus, we have 7^2*3^3=1323.

a^2*6^3=1323, 

a=6.125

agree with ElectricPavlov !

Rewrite the quadratic equation as a(x-h)^2+k, where h and k are parts of the vertex.

y=a(x-2)^2-4

Since (4,12) is a point on the parabola, we can substitute these values in the equation.

12=a(4-2)^2-4, 12=4a-4, 4a=16, a=4

Now, the equation is complete: \(y=4(x-2)^2-4, y=4x^2-16x+12\)

We then plug y=0, to get x=3, x=1.

Thus, m-n=3-1=2.

Each coin has two sides, heads or tails, so the total number of possible outcomes is \(2^{10}=1024\).

The number of ways of getting heads is 10 C 6   *   10  C  7   + 10   C  8  +  10 C  9   + 10  C  10=210+120+45+10+1=386 ways.

193/512  but I might be worng

Since all four sides in a rhombus are equal, each side should be equal to \(\frac{68}{4}=17.\)

The key idea in this problem is that the two diagonals in the rhombus are perpendicular to each other, and most importantly, the diagonals bisect each other...

We have a right triangle, with one leg as eight(8) units and the hypotenuse has length seventeen(17) units.

By the Pythagorean theorem, the other side length has length fifteen(15) units.

Thus, the second diagonal has length \(2*15=30\) units.

We have \(\frac{x}{y^3}\) and \(x\sqrt{z}\) are the two restrictions...

\(3\sqrt{12}=y*\sqrt{75}, y=\frac{6}{5}=1.2.\)