hairyberry

Volume = base * height / 3.

Our base area is 4 * 4 = 16.

Therefore, the volume is 16 * 6 / 3 = 32.

Here:

https://web2.0calc.com/questions/coordinates_61733#r1

f(x - 3) represents a translation right by 3.

Therefore, the whole graph is shifted right by 3, meaning the vertex shifts right by 3.

Shifting right by 3 means that 3 is added to the x coordinate.

Therefore the new vertex is (1 + 3, -2) = (4, -2).

Alternatively, if you don't know translations: Use vertex form:

\(f(x)=a(x-1)^2-2\) represents a parabola of vertex (1, -2), where a is a constant

\(f(x-3) = a((x-3)-1)^2-2=a(x-4)^2-2\), so looking at this form, you know the vertex is (4, -2).

\(3mn-30m+6n+c\)

\(3m(n - 10) + 6n + c\)

\(3m(n-10)+6(n-10)+60+c\)

\((3m+6)(n-10)+60+c\)

\(c+60 = 0\)

\(c=-60\)

The constant is -60.

\(a_n = a_{n - 1} + a_{n - 2}\)

\(a_{n-2}=a_n-a_{n-1}\)

\(a_9=4-1=3\)

\(a_8=1-3=-2\)

\(a_7=3-(-2)=5\)

\(a_6=-2-5=-7\).

Draw a rectangle in the middle like so:

The width is r-13, while the length is r - 10, and the diagonal is r:

Therefore, use the pythagorean theorem.

\({(r-10)}^{2}+{(r-13)}^{2}={r}^{2}\)

\(r=\frac{46\pm \sqrt{1040}}{2}=23 \pm \sqrt{260} = 23\pm 2\sqrt{65}\)

We know because the distance of 13 to the wall, r > 13, therefore the only solution is \(23+2\sqrt{65}\).

The lowest prime that satisfies this is 5, because 5 is right after 4.

To find the highest, do 200/5 = 40, and round down which is 37.

There are a total of 10 primes between these 2:

5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Lets count the total number (excluding those that have already been counted)

Groups with 5: 7, 11, 13, 17, 19, 23, 29, 31, 37 total : 9

Groups with 7: 11, 13, 17, 19, 23 total : 5

Groups with 11: 13, 17 total : 2

No more after this ~

There are a total of 9 + 5 + 2 = 16 total products.

Lets use complementary counting, it is much each to count all the cases where the freshman don't stand next to each other.

There are 7! ways to arrange 7 distinct people.

To have no two freshman stand next to each other, we must start with a freshman and alternate between freshman and sophomore like this:

F S F S F S F.

We can arrange the freshman in this arrangement 4! ways, and the sophomores 3! ways, so there are \(4!*3!\) ways for this arrangement.

Therefore there are 7!-3!*4! = 4896 ways to do this.

The word "ARCH" has to contain the letter A.

There is only one letter A, therefore it much start there.

We can go hoizontally, diagonally, or vertically all once.

Therefore there are 3 ways to do this.