BuilderBoi

Sorry, I didn't do the best job... Hopefully this is better!

There will be some answer for \(k\), but there will also be another answer, \(24 - k\).

This is because \({n \choose k} = {n! \over k!(n-k)!}\). However, \({n \choose n - k} = {n! \over (n-k)!k!}\). For example, \({5 \choose 2} = {5! \over 2! 3!} = 10\), and likewise, \({5 \choose 3}=10\).

The sum of 2 and 3, which are \(k\), and \(n - k\) ,respectively, both sum to \(n\), or 5. 

Likewise, applying the same logic here, we find that the sum will be \(\color{brown}\boxed{24}\), because \(n = 24\)

Think about all the distinct integer factor pairs of 100: 

1 and 100

2 and 50

4 and 25

5 and 20 

Which has the smallest sum?

The amount of time that would have passed once the ball reached 30 feet can be written by subsituting 30 in for y. 

This gives: \(30 = -16t^2 + 60t\). Subtracting 30 from both sides gives us this quadratic: \(0 = -16t^2 + 60t - 30 \) 

We can solve for t using the quadratic formula, which is \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), because the quadratic is in the form \(0 = ax^2 + bx + c\)

Can you take it from here?

Guest: I'm so glad I could help!

Melody: Thank you!

Also, I just realized my original answer has a typo, it should be "For a quadratic to have repeated roots", not "For a quadratic to have distinct roots"

Yes, it is. 

Let the starting term be \(a\) and let the common difference be \(d\).

The \(n_{\text {th}}\) term of this sequence is \(a + d(n-1)\). This means that the starting (1st) term is \(a\), and the second term is \(a + d\)

Thus, we can form the following system: \(a+(a+2d)+(a+4d) = -12\) and \(a(a+2d)(a+4d) = 80\)

Now, we just have to solve for the 10th term, or \(a + 9d\).

Can you take it from here?

We will start with the numerator of the square root. 

For the first fraction(\(\large{5 \over {\sqrt {20}}}\)), we can rationalize the denominator, by multiplying the fraction by \(\large{\sqrt {20 } \over {\sqrt{20}}}\). This gives us \(\large{{5 \sqrt {20 }} \over 20}\)

We can rewrite \(\sqrt{20}\) as \(2 \sqrt 5 \). This means we have \(\large{5 \times 2 \sqrt 5 \over 20}\), which simplifies to \(\large{\sqrt 5 \over 2}\)

Repeating this process to the second fraction (\(\large{\sqrt{845} \over 9}\)) gives us \(\large{{13 \sqrt 5} \over 9}\). 

Adding all the terms up gives us: \(\large{{{26 \sqrt5}\over 18} + {18\sqrt5 \over 18} + {9 \sqrt 5 \over 18}} = {53 \sqrt 5 \over 18}\).

This means that we have: \(\large{\sqrt{{53 \sqrt5 \over 18} \over \sqrt 5}}\). 

Now, we just have to cancel out the common terms, and we have our answer. 

Can you take it from here?

Consider it as 2 seperate series: \(1 + 2 + 3 + 4 + \cdot \cdot \cdot+ 24 +25\)and \(4+8+12+16 + \cdot \cdot \cdot + 96+100\)

The sum of the first series is simple, as the sum can be defined as: \(25 \times 26 \div 2 = 325\)

To make things easier, we can express the second series in multiples of 4 as follows: \(1(4) + 2(4) + 3(4) + 4(4) + \cdot \cdot \cdot + 24(4) + 25(4)\)

This can be rewritten as: \((1 + 2 +3+4\cdot \cdot \cdot +24 +25)\times 4 \). Thus, the sum of the second series is: \(325 \times 4 = 1300\)

Can you find the answer from here?

We can simplify the given expression to: \(a = 105 (\mod 11)\)

This means that the remainder of a when divided by 11 is 105. 

However, we can simplify this furthur, because 105 can "go" into 11 9 times, with a final remainder of 6. 

This means that \(a = 6 (\mod 11)\). 

Can you find the remainder from here?

Note that \({24 \choose k} = {24 \choose 24 - k}\)

This means that the sum of all integers \(k\), that work, is \(\color{brown}\boxed{24}\)

For a quadratic to have distinct roots, it's must be the form: \(a^2x^2 \pm 2abx + b^2\)

For example, applying this to the fifth one yields: \(1^2 x^2 - 2 \times 1 \times 14x + 14^2\), which equals: \(x^2- 28x+196\), this, isn't true, so it isn't 5. 

Can you do the rest?

Let \(a\) and \(b\) be the legs of the triangle, and let \(c\) be the hypotenuse. From the Pythagorean Theorem, along with the given information, we have the following system of equations: 

\(a^2 + b^2 = c^2\)                   (1)

\(a+b+c=26\)               (2)

\(a^2+b^2+c^2=288\)       (3)

Subsituting (1) in (3) gives us: \(2c^2 =288\). Solving this equations shows that \(c = 12 \)

Now, we have the system:

\(a^2 + b^2 = 144\)         (1)

\(a+b = 14\)                (2)

Recall the identity: \(a^2 + b^2 = (a+b)^2-2ab\)

This means we have the equation: \((a+b)^2-2ab=144\)

Now subsituting in (2) gives us: \(14^2 - 2ab = 144\)

Can you solve for the area from here?

Hint: the area, because it is a right triangle, is \({ab \over 2}\)