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x^2 - 5x + 4 = 0

x^2 - x + 4 - 4x = 0

x(x-1) - 4(x-1) = 0

(x-4)(x-1)=0

x=4, x=1

9(x+3) = -8 + x

Expand brackets

9x + 27 = -8 + x

Collect like terms

8x = -35

x = -35/8

2x/(x+2) = x/(x-4)

Get rid of the divisors

2x(x-4) = x(x+2)

Expand

2x^2 - 8x = x^2 +2x

Collecting like terms

x^2 - 10x = 0

Factoring

x(x-10) = 0

x=0, x=10

We simplify the expression by simplifying each radical term separately. [ \begin{aligned} \frac{1}{\sqrt{36}} - \sqrt{27} - \frac{1}{\sqrt{27}} - \sqrt{18} + \frac{1}{\sqrt{18}} - \sqrt{9} =& \frac{1}{6} - 3\sqrt{3} - \frac{1}{3\sqrt{3}} - 3\sqrt{2} + \frac{1}{3\sqrt{2}} - 3\ =& \frac{1}{6} - 3(\sqrt{3} + \sqrt{2}) + \frac{1}{3}(\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}}) - 3\ =& \frac{1}{6} - 3(\sqrt{3} + \sqrt{2}) + \frac{1}{3}(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2}\sqrt{3}}) - 3\ =& \frac{1}{6} - 3(\sqrt{3} + \sqrt{2}) + \frac{\sqrt{2} + \sqrt{3}}{3} - 3\ =& \frac{1}{6} + \frac{\sqrt{2} + \sqrt{3}}{3} - 3 - 3(\sqrt{3} + \sqrt{2})\ =& \boxed{\frac{1}{6} + \frac{\sqrt{2} + \sqrt{3}}{3} - 6(\sqrt{3} + \sqrt{2})} \end{aligned} ]

We can simplify this expression by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of the denominator is (2+i)(3−i)=6+5i−3i−i2=6+2i−1=5+2i​.

For the system to have infinitely many solutions, the left-hand sides must be equal to each other, regardless of the values of a and b. Thus, 3a + 2b = 6a + 2b, which implies a = 0. Substituting this value into the first equation yields 2b = 2. Hence, b = 1. Then, the second equation becomes 0 - mb = k - 3(0) - m(1), which simplifies to k = m.

Therefore, k = m = \boxed{1}.

The side lengths of the original square and the new square are 3 and 5. Therefore, there are two values of k: 5/3 and -5/3. Note: you can dilate by a negative scale factor, think about the square on a cartesian plane. Therefore sum of the values of k are 0.

Combining like terms, we get the equation:

\(5x^2+4x-6=0\)

Vieta's formulas states that the sum of roots for a quadratic ax^2+bx+c is -b/a. Therefore, the sum of the roots for this equation is -4/5. Since the prom is asking for a+b-8, the answer is -44/5.

A quadratic whose roots are 1 and -1 in factored form looks like this:

\((x-1)(x+1)\)

Expanding this out, and it becomes \(x^2+1\)

Now we fill in the blanks; the first blank is 0, since there is no "x" term. The second blank is 1.

If all three resulting pieces are shorter than 2 units, then the sum of the 3 pieces would be less than six, the length of our big stick. Therefore, there are no ways to make 3 pieces shorter than 2 units, so the probability is 0.

\((11 - 7i)*(2 + i) = 22+11i-14i+7=\color{blue}29-3i\)

\((11 - 7i)*(2 + i) = 22+11i-14i+7=\color{blue}29-3i\)

To make writing easier, replace √5 

by x in the left side of the original expression. After replacing, you will get this polynomial (3+x)*3*(5+3x))^3.
It is

(3 + x)*3*(5 + 3x))^3 = 81x^4 + 648x^3 + 1890x^2 + 2400x + 1125. (*)

Now, to get first integer number of the right side, I consider the terms of this polynomial (*) with even degrees

even(x) = 81x^4 + 1890x^2 + 1125

and substitute there x² = 5.

I get then "second integer number in the answer" = 648*5 + 2400 = which is easy to compute = 5640. Therefore, the final ANSWER is 12600 + 5640* √5 (**)

Finally, to check the answer, I calculated (**) and compared with the left side of the original expression. I got the same number, which confirms correctness of my solution.

(LaTeX version)

\(\textbf{Compute}\)

\(\binom{5}{3} - \binom{6}{3} + \binom{7}{3} - \dots - \binom{22}{3} + \binom{23}{3} - \binom{24}{3} \)

(LaTeX version)

\(\textbf{Fill in the blanks to make the equation true. (Each blank should contain a 2-digit number.) }\)

\(18\times18 + 19\times18 + 20\times18 + \dots+ 45\times18 =\boxed{\phantom{X}} \times \boxed{\phantom{X}} \)

To solve this problem, we can use the principle of inclusion and exclusion.

Let A be the set of all ways to distribute 8 indistinguishable balls among 15 distinguishable boxes.

Let B_i be the set of all ways to distribute the balls such that box i is empty.

We want to count the number of ways to distribute the balls such that at least 3 boxes are empty. This is the same as counting the number of ways to distribute the balls such that 0, 1, or 2 boxes are empty.

We can use the following formula:

|A - B_1 - B_2 - ... - B_n| = |A| - |B_1| - |B_2| - ... - |B_n| + |B_1 \cap B_2| + |B_1 \cap B_3| + ... + |B_{n-1} \cap B_n| - |B_1 \cap B_2 \cap B_3| - ... - |B_{n-2} \cap B_{n-1} \cap B_n| + ... + |B_1 \cap B_2 \cap B_3 \cap ... \cap B_n|

where |A| is the cardinality of set A, and |B_i| is the cardinality of set B_i.

In our case, we have:

|A| = \binom{15}{8} = 6435

|B_i| = \binom{14}{7} = 3432

|B_1 \cap B_2| = \binom{13}{6} = 1771

|B_1 \cap B_3| = |B_2 \cap B_3| = \binom{12}{5} = 792

|B_1 \cap B_2 \cap B_3| = \binom{11}{4} = 330

Therefore, the number of ways to distribute the balls such that at least 3 boxes are empty is:

6435 - 3 * 3432 + 3 * 1771 - 3 * 792 + 1 * 330 = \boxed{1818}

Does that include the 4-digit integers, 6-digit etc.?

We have 8 slots for the digits of our number, and we need to place a 1 in 4 of those slots. This can be done in C(8,4) ways.

Once we have placed the 1's, we can fill in the remaining slots with 3's. This can be done in 4! ways.

Therefore, the total number of positive integers that satisfy both of the given conditions is C(8,4)*4! = 416.