BuilderBoi

Note that \(\angle B = 3 \times \angle QBP = 3 \times 12^\circ = 36^ \circ\)

This means that \(\angle C = 180 - 36 - 35 = 109\)

So, \(\angle PCB = {109^ \circ \over 3} = 33^ \circ\). 

But, recall that \(\angle PBC = \angle QBP = 12\)

This means that \(\angle BPC = 180 - \angle PBC - \angle QBP = 180 - 33 - 12 = \color{brown}\boxed{135^\circ}\)

The probability is the shaded region over the total square: 

Disregard this answer.....

REVISED ANSWER: There are \(9^4 = 6561\) ways to get the numbers. 

The easiest way to solve this is to use complementary counting, in which we count what we don't want, and subtract that from 1

Note that the probability of getting an odd number is \({5 \over 9}^4 = {625 \over 6561}\)

The probability of getting a 2 or 6 is \({2 \times 5^3 \times {4 \choose 3} \over 6561} = {1000 \over 6561}\) note that 4 and 8 make the product divisible by 4. 

So, the total probability is \(1 - ({625 \over 6561} + {1000 \over 6561} ) = \color{brown}\boxed{4936 \over 6561}\)

LOL, Thank you!!

Note that\(BE = B'E = 12\)

By the Pythagorean Theorem, \(AB' = \sqrt{B'E^2 - AE^2} = \sqrt{144 - 64} = \sqrt{80} = 4 \sqrt 5\)

Also, note that by the Pythagorean Theorem, \(B'F = DF^2 + B'D^2 = CF^2 + BC^2\)

SImplifying the equation gives us: \(4 + BC^2 = 324 + B'D^2\)

But, note that \({B'D} = {BC - AB'} = BC - 4 \sqrt 5\)

This means that our equation simplifies to \(4 + BC^2 = 324 + (BC - 4 \sqrt5)^2\)

Now, just solve for \(BC\), and I assume you can take it from here?

The graph of a quadratic can be written as \(y = a(x - h)^2 + k\), and \((h, k)\) is the vertex.

Now, plug in the vertex, to get this: \(y = a(x + 4)^2\)

Now, plug in the point \((1, -15)\) and solve for a. 

Proofread your question carefully 

Write an equation: \(4200 > t(70 - t)\)

Now, just solve for t...

\({11 \over 30} \times {10 \over 29} \times {12 \over 28} \times {11 \over 27} \times {10 \over 26} \times {5 \choose 2} = \text{____}\)