BuilderBoi

We begin by cross-multiplying: \(2x(x+4) = (x-5)(x+5)\)

We can expand both sides to get: \(2x^2+8x=x^2-25\)

We can move everything to the left-hand side, which gives us: \(x^2+8x+25=0\)

We now have to solve for x using the quadratic formula. 

Can you do it?

Hint: The answer isn't real.

We can rewrite this as: \(2^{2^{10}} \times 2^{3^{20}} \times 2^{4^{30}}\)

This can be rewritten as: \(2^{20} \times 2^{60} \times 2^{120}\)

Can you do it from here?

Remember that we add the exponents when we multiply. 

Here is a diagram that will help: 

The area of \(\triangle ABC\) is \([ACDE] - [AEB] - [CDB]\) 

Can you solve it now?

Draw the line \(x = 1\) and let the red line be \(y = kx\). 

The points of intersection are: \((1,4)\), \((1, 2)\), and \((1, k)\) .

Using the Pythagorean Theorem, we find that the lines of the triangle bounded by the 2 lines and x= 1 are \(2\), \(\sqrt 5 \), and \(\sqrt {18}\).

Let the distance between \((1,k)\) and \((1,4)\) be \(d\). This means the distance from the points \((1,2)\) and \((1, k)\) is \(2 - d\).

Using the Angle Bisector Theorem, we can form the following equation: \({ d \over \sqrt{18}} = {2 - d \over \sqrt{5}}\)

Now, we have to solve for d and subtract that from 4.

Can you take it from here? 

\(6 \over 19\) chance for the first marble

\(5 \over 19\) chance for the second marble

\(8 \over 19 \) chance for the third marble. 

Because they all "rely" on each other, the probability is the product of all 3 cases. 

Aww... Thank you!!!

Yes, that is correct!!!

The perimeter of the trapezoid is the sum of the sides + the sum of the line cutting the square. 

The length of the sides (not including the line) is \(4a\). One of the sides is 2a (labeled a in the diagram), and the other two sides (b and c in the diagram) will always add to 2a. 

We can then use the Pythagorean Theorem for the length of the bisecting line, and we find that its length is \(a \sqrt 5 \). 

Thus, the perimeter of the shape is \(a(4 + \sqrt 5)\). Can you do the rest?

Diagram:

We have the system: 

\(\sqrt{xy} = 21\)

\({{x+y} \over 2} = 75\)

We can simplify both equations to: 

\(xy= 441\)                (1)

\(x+y=150\)           (2)

Now, we just have to solve this system. Can you do it?

Because the circle has a radius of 1, we know the rectangle's dimensions are \(4 \times 2 \). 

Note that the perimeter of the shaded region is the perimeter of the rectangle plus the perimeter of the 2 circles. 

The circumference/perimeter of the circle can be found using the formula \(2 \pi r\). 

Can you take it from here?