Capban

Let the missing number be k^2

then 18x = 2xk

and so k = 9

this makes the missing number be 9^2 = 81

\(\frac{t}{3} - 5 < t - 2 + 2t \le -3t + 7\)

\(t - 15 < 3t - 6 + 6t \le -9t + 21\)

\(t - 15 < 9t - 6 \le -9t + 21\)

\(-15 < 8t - 6 \le -10t + 21\)

\(8t - 6 \le -10t + 21\)

\(18t \le 27\)

\(t \le \frac{3}{2}\)

\(-15 < 8t - 6\)

\(-9 < 8t\)

\(t > -\frac{9}{8}\)

\(\boxed{t = (-\frac{9}{8}, \frac{3}{2}]}\)

If we add all the equations, we get \(10a + 10b + 10c + 10d = 0\).

We then divide by 10 to get \(a + b + c + d = \boxed{0}.\)

If A's statement is true then it means that C's statement was fake, which means C did break it, but A's statement said that D broke it, and C and D can't both broke the vase, so A's statement is fake;

If B's statement is true then E broke the vase, but C's statment would be fake and make C the one that broke the vase, so B's statement is fake; 

If C's statement is true then D's statement would be fake, so A was telling the truth when he said D broke it, but A's statement is fake, and so D couldn't have broke the vase, so C's statement is fake;

If D's statement is true then A's statement was fake, and that D didn't break the vase. This also means that B and E vouched each other, therefore they didn't break the vase. This statement doesn't cause any error, so C broke the vase. But just to be safe, let's also check E's statement.

E's statement is the same as B's statement, so E's statement is also fake.

Therefore, D's statement is true and C broke the vase.

We know that shortleg^2 + longleg^2 = \(\sqrt{5}^2\) or 5

If the shorter leg is x then the longer leg would be 3x

Therefore we have:

\(x^2 + 9x^2 = 5\)

simplify it to

\(10x^2 = 5\)

divide 4 each side

\(x^2 = \frac{5}{10}\)

square root both side

\(x = \sqrt{\frac{5}{10}}\)

So the shorter leg is \(\sqrt{\frac{5}{10}}\) and the longer leg is \(3 \sqrt{\frac{5}{10}}\)

Now we need to find the area

Multiply them:

\(3\sqrt{\frac{5}{10}}\cdot \sqrt{\frac{5}{10}} = 3 \cdot \frac{5}{10} = \frac{15}{10} = \frac{3}{2}\)

And don't forget we also have to divide by 2

\(\frac{3}{2} \div 2 = \frac{3}{4}\)

For all 30° 60° 90° right triangles if x = the shortest side's length the hypotenuse would be 2x

So the hypotenuse of the first triangle would be \(2 \times 1 = 2\) 

We could see that the second triangle is also 30° 60° 90°

2 is the length of the shortest side in the second triangle so \(x = 2 \times 2\)

\(x = 4\)

We are trying to find n

We could make a right triangle and use the Pythagorean equation \(a^2+b^2=c^2\) to find out the hypotenuse in this case is n

For this \(a\) will be 8 and \(b\) will be 12

\(8^2+12^2=c^2\)

\(64+144=c^2\)

\(208 = c^2\)

\(c = \sqrt{208}\)

5m-3 > 8m+21 + 7m+3

5m-3 > 15m + 24

5m > 15m +27

-10m > 27

10m < -27

m < \(-\frac{27}{10}\)

m < \(-2\frac{7}{10}\)

The greatest integer value for m would be -3.

There are 10 ways to pick a project manager

\(\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36\) ways to pick two deputy project managers

\(\binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21\)ways to pick five engineers

\(10 \times 36 \times 21 = \boxed{7560}\)

John can use \(\frac{1}{7}\)of his money to buy 2 rulers and 4 pens.

John can also \(1 - \frac{2}{7} = \frac{5}{7}\)of his money to buy 6 rulers, 7 books, and 12 pens.

We could also see that he can buy 6 rulers and 12 pens using \(\frac{1}{7} \times 3 = \frac{3}{7}\)of his money.

7 books cost \(\frac{5}{7} - \frac{3}{7} = \frac{2}{7}\)of his money.

\(7 \times 3 = \boxed{21}\)

4x = 3y

x = \(\frac{3}{4}y\) 

\(\frac{2x+y}{3x+4y}\) 

\(\frac{\frac{6}{4}y + y}{\frac{9}{4}y + 4y}\) 

 \(\frac{\frac{10}{4}y}{\frac{25}{4}y}\) 

\(\frac{10y}{25y}\) 

\(\frac{2y}{5y}\) 

\(\boxed{\frac{2}{5}}\)

x(2x-7)=8

2x^2 - 7x = 8

2x^2 - 7x - 8 = 0

Using the quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) 

\(\frac{7 \pm \sqrt{49+64}}{4}\) 

\(\frac{7 \pm \sqrt{113}}{4}\) 

\(\frac{7 + \sqrt{113}}{4}\) and \(\frac{7 - \sqrt{113}}{4}\) 

So m = 7 n = 113 p = 4

7 + 113 + 4 = \(\boxed{124} \)

p(1,-1) = 1 - (-2) = 1 + 2 = 3

p(-5, -2) = -5 - (-4) = -5 + 4 = -1

p(3, -1) = 3 - (-2) = 3 + 2 = \(\boxed{5}\)

We could use Complementary Counting for this.

The probability that it will not slow at all is      \(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}\)        

Then we minus the total by   \(\frac{1}{27}\) 

\(1 - \frac{1}{27} = \frac{26}{27}\)