Maxematics

11012_9 = 1 * 6561 + 1 * 729 + 0 * 81 + 1 * 9 + 2 * 1 = 6561 + 729 + 9 + 2 = 7301

For a number to be written as \(0.\overline{abcdef}\), you must be able to write it as \(\frac{abcdef}{999999}\). Therfore, we can just divide 999999 by 17 to find the factor we need to multiply 2/17 by. Using a bit of scrap paper, we find that 999999 / 17 is not a whole number (it's aapproximately 58823.471). Therefore, it is impossible to make a 6-digit number that has that property.

You can arrange the cities into a straight line like this. Therefore, the distance from Capital City to Yourtown is 1 mile.

Let's find the rate for each person. You can do this by using the formula "work = (rate)(time)(number of people). Plugging in we get: 

\(1000 = r(10)(10)\)

\(\frac{1000}{100}=r\)

\(r = 10\)

Now we can use the rate to find out the time in the second equation.

\(1000 = 10(t)(5)\)

\(\frac{1000}{50} = t\)

\(\mathbf{t = 20}\)

Therefore, it would take 20 minutes for 5 members to paint a 1000 square foot wall.

Let's write the number of liters in Bucket A as "a", he amount in Bucket B "b", and the amount in Bucket C "c". We can now write the following equations:

\(a + b + c = 17\)

\(a = 3b\)

\(c = b - 2.5\)

We can now use subsitution to solve for each of them.

\(3b + b + b - 2.5 = 17\)

\(5b = 19.5\)

\(\mathbf{b = 3.9}\)

So Bucket B has 3.9 liters.

For a polynomial with degree two to have one solution, you must be able to write it as (x + y)^2. We can rewite the equations like this:

\((2x+7)(x-4)=jx-31\)

\(2x^2+7x-8x-28=jx-31\)

\(2x^2-x-jx+3=0\)

\(2x^2-(j+1)x+3=0\)

Now we must make an equation in the form of x^2 + 2xy + y^2 or x2 - 2xy + y^2

\(\sqrt{2}^2x^2 - (j+1)x+\sqrt{3}^2=0\)

Therefore, x = sqrt(2) and y = sqrt(3). Now we see that j+1 = 2(sqrt(2))(sqrt(3)) or -2(sqrt(2))(sqrt(3)). We can now sove for both values of j.

\(j+1 = -2\sqrt{2}\sqrt{3}\)

\(\mathbf{j = -1 - 2\sqrt{6}}\)

\(j+1 = 2\sqrt{2}\sqrt{3}\)

\(\mathbf{j = 2\sqrt{6}-1}\)

So the only two values of j that make the equation have one solution are 2sqrt(6)-1 and -2sqrt(6)-1

Since all of he numbers in the set are divisible by 2, we can divide them all by 2 and we would still have the same number of unique sums (I can easily prove this if you need me to). So now our set is {2, 3, 4, 5, 6,...,77}. The smallest possible sum in this set is 2 + 3 + 4 = 9, and the largest sum is 75 + 76 + 77 = 228. Because all other numbers in the set are one more then the one before it, we can easily add one to any sum by just increasing a number by one. This means that you can only make the sums of all the numbers from 9 to 228. There are 220 numbers in that range, so there are 220 unique sums that you can make.

We can use the Quadratic formula to find the value of v. 

a = 5, b = 21, c = v

\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(\frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\)

\(\frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\)

\(\sqrt{201} = \sqrt{441-20v}\)

\(201 = 441-20v\)

\(20v = 240\)

\(\mathbf{v=12}\)

Therefore, the value of v is 12

Line EF is parallel to line BC, due to both being lines of the square. Therefore, angle FEA is equal to angle ACB and angle AFE is equal to angle ABC. We can now prove that triangle AFE is similar to triangle ABC due to AA similarity. Let's suppose that the side length of the square is x. Using the ratios of the similar triangles:

\(\frac{3-x}{x} = \frac{3}{1}\)

\(3-x = 3x\)

\(4x=3\)

\(x=\frac{3}{4}\)

So the side length of the square is 3/4. Therefore, the area of the square is (3/4)^2 = 9/16 square units.