BuilderBoi

Rewrite the equation using the known values of x and y in terms of k as: \(k+1 = 4k - 14\)

Solving for k, we find \(k = 5\)

Substituting the known value of k into the point, we find it passes through the point \(\color{brown}\boxed{(5,6)}\)

There are 2 options for 1 question

There are another 2 options for the other question

The remaining 8 must all be false.

There are \(10 \choose 8 \) ways to choose which questions are guaranteed to be false

So, in total, there is: \(2 \times 2 \times {10 \choose 8} = \color{brown}\boxed{180}\) ways to answer. 

1) \(\color{brown}\boxed{f(x)= 63-7(x-1)}\), where \(1 \leq x \leq 6\)

2) There are 6 rows, the bottom row with 63 and the top row with 28. 

Writing as factors of 7, we have: \((7 \times 4) + (7 \times 5) + \space ... \space + (7 \times 9)\)

We can rewrite this as: \(7 \times (4+5 + \space ... + \space 9)\)

This is equal to \(7 \times 39 = \color{brown}\boxed{273}\)

Reflect point \((18,0)\) over the axis of symmetry, which is \(x=12\).

After the reflection, the other x-intercept is at \(\color{brown}\boxed{(6,0)}\)

There are 4 corners to put them in.

There are 7 dots in 3 rounds, so there is \(4 \times 7 \times 7 \times 7 = \color{brown}\boxed{1,372}\)

lol, at least we have the same answers :)

We have 6 equations: 

\(a+b=10\)    (1)

\(a+c=18\)    (2)

\(a+d=19\)    (3)

\(b+c=22\)     (4)

\(b+d=23\)    (5)

\(c+d=31\)    (6)

Using equations 2 and 3, we see that \(d = 1+c\)

Subbing this into (6), we get: \(c+c+1=31\), meaning \(c=15\)

Subbing this into (2), we find \(a=3\)

Subbing the value of c into (4), we find \(b=7\)

Subbing the value of c into (6), we find \(d = 16\)

Thus, the values are \(\color{brown}\boxed{3,7,15,16}\)

I solved this problem using co-ordinate geometry. 

Let the bottom-right corner be the origin. The equations for lines \(AN\) and \(DM\) are \(y = {1 \over 2}x\) and \(y=-2x+1\)

Letting these 2 equations equal each other, we have: \(0.5x=-2x+1\).

Solving, we find \(x = 0.4\)

Thus, the area of \(\triangle AMO\) is \(1 \times {2 \over 5} \div 2 = \color{brown}\boxed{1 \over 5}\)

Here is the diagram:

Here's another way: Draw altitude \(AM\).

\(\triangle {AMC} \) is a 30-60-90 triangle, meaning \(CM = 5\) and \(AM = 5 \sqrt3\)

Now, \(\triangle {AMB}\) is a right triangle, so applying the Pythagorean Theorem to \(\triangle AMB\), we find that \(MB = \sqrt{69}\)

Since \(CB = CM + MB\), \(CB = \color{brown}\boxed{5 + \sqrt{69}}\)

The formula for the volume of a cone is: \({r^2 \pi h } \over 3\)

Plugging in the values, we have: \({18^2 \pi 20 \over3} = \color{brown}\boxed{2160 \pi}\)