WhyamIdoingthis

Note that k CANNOT be equal to zero.

One root is \(x = {-7 + \sqrt{49-12k} \over 6}\), while the other is \(x = {-7 - \sqrt{49-12k} \over 6}\).

Their reciprocals are \(6 \over -7+\sqrt{49-12k}\) and \(6 \over -7-\sqrt{49-12k}\).

Adding those gives this nice expression: - \(7 \over k\) 

This is equal to one.

-7/k = 1

k = -7

I assume that the buses are all the same size.

259 - 27 = 232

232/8 = 29

There were 29 kids in each bus. 

Wedge.

Cube both sides: 

3 - x = -27/8

\(x = 3 + {27\over8}\)

\(x={51\over 8}\)

Note that f(x) will always put out a value greater than or equal to four for an imput of less than or equal to four. 

For x<=4:

f(f(x)) = a(f(x))+b = a(8 - x) + b = x

We want (a,b) such that a(8 - x) + b = x for x<=4.

-ax + 8a + b = x

There is only one term on the left that contains x, which means that it must be equal to x. The only value of a that would do that is a = -1.

x - 8 + b = x

b = 8

a + b = 7

(I didn't check this, it could be wrong)

The sum of the exterior angles is always 360 degrees. This means that the exterior angles of this triangle are 90 degrees, 135 degrees, and 135 degrees. This would make the interior angles 90 degrees, 45 degrees, and 45 degrees, which makes the smallest angle 45 degrees.

I believe if you graph it on a graphing application of some sort the answer should be pretty clearly \((-\infty, \infty)\).

This is equal to (1-x²)¹⁰, since (1+x)^10 x (1-x)^10 = ((1+x)(1-x))^10.

A binomial raised to any power n has n+1 terms (assuming nothing simplifies).

There will be 10+1= 11 terms upon combining all like terms (I think).

Recall the geometric sequence formula for the nth term: \(a_n = a_1(r)^{n-1}\)

\(a_1 = 9\)

r = 1/6

n = 10

\(a_{10} = 9({1 \over 6})^9\)\(= {9 \over 6 ^ 9}\)\(= {1 \over 2^2 \cdot 6^7}\)

If you need a decimal form I believe that the calculator on this website should be able to provide you with that.

The first term is 3/2⁰.

The second one is 3/2¹.

The third is 3/2².

The pattern continues, because it is a geometric sequence. The eighth term would be 

3/2⁷ = \(3 \over 128\)