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Using the hint at the very end, we can multiply $5.5x+7.5y=930$ by -2 to make it easier:

$-11x-15y=-1860\\ 12x+15y=1920$

Add the equations together to cancel out the y term:

$x=60$

Now that we know x, we can plug x back into any of the 2 equations above to solve for y (I chose the second one because it seems simpler):

$12(60)+15y=1920\\ 720+15y=1920\\ 15y=1200\\ y=80$

Therefore, the answer is:

$\boxed{x=60, y=80}$

According to the rational root theorem, the possible rational roots are the possibilities of the factors of the constant term divided by factors of the coefficient of the highest degree term.

Therefore, the possible roots are $\pm1,\pm2,\pm3,\pm6$.

You could test out these individual values and plug them into the equation.

The ones that work are -3, -1, and 2.

Therefore the factored form of this polynomial is (x+3)(x+1)(x-2)

This can be simplified to:

$3^{-2}=x+2$

Then, just solve the linear equation:

$\frac{1}{9}=x+2\\ x=\boxed{-\frac{17}{9}}$

I'm not entirely sure, but I think that the largest possible length of the shortest side occurs when it's a rhombus.

Therefore, every side will be sqrt(3^2+4^2)=5, so 5's our answer

We can use complementary counting for this problem by calculating the probability that it will not snow at all first.

That is equal to 

$(\frac{1}{2})^3\cdot(\frac{2}{3})^4=\frac{1}{8}\cdot\frac{16}{81}=\frac{2}{81}$

But we're trying to find the probability that it will snow at least once, so the answer is $1-\frac{2}{81}=\frac{79}{81}$

Here's a very similar problem: https://web2.0calc.com/questions/please-help-complementry-probabilty-problem

Notice that for 2 numbers a and b, $(a+1)(b+1)=ab+a+b+1$ This is very similar to the equations above, so we can rewrite the equations like this:

$(x+1)(y+1)-1=19 \rightarrow (x+1)(y+1)=20\\(y+1)(z+1)-1=29\rightarrow(y+1)(z+1)=30\\(z+1)(x+1)-1=5\rightarrow(z+1)(x+1)=6$

Let us replace (x+1) with a, (y+1) with b, and (z+1) with c, so it is easier to solve:

$ab=20\\bc=30\\ca=6$

Multiply all the equations and then take the square root:

$a^2b^2c^2=3600\\ abc=60$

Now, to solve for a, b, and c, just divide abc by one of the equations of ab, bc, or ca:

$a = \frac{abc}{bc}=2\\ b = \frac{abc}{ac}=10\\ c = \frac{abc}{ab}=3$

Now, subtract each of a, b, and c by 1 to solve for x, y and z, respectively:

$x =1\\ y=9\\ z=2$

The difference between 2 numbers in this set will always be even because an odd number minus an odd number will always be even.

The smallest attainable value is 2, and the biggest attainable value is 18. Every value between these 2 numbers will also be attainable.

That means the answer to this question is: there are 9 distinct positive integers that can be represented as the difference of 2 numbers in that set.

(they are 2, 4, 6, 8, 10, 12, 14, 16, and 18)

The smallest possible square that contains at least 12 black squares are the 5 by 5 squares; they either contain 12 or 13 squares. Therefore, we just need to find the number of 5 by 5, 6 by 6, 7 by 7, and 8 by 8 squares, since they all contain at least 12 squares.

Number of 8 by 8 squares = 1^2 = 1

Number of 7 by 7 squares = 2^2 = 4

Number of 6 by 6 squares = 3^2 = 9

Number of 5 by 5 squares = 4^2 = 16

Therefore, the total number of squares with at least 12 squares is 1+4+9+16 = 30  

Good answer, but I think you misread the problem. It said that the bigger angle should be 5 times the measure of the smaller angle. Therefore,

x+5x=180

6x=180

x=30