$\displaystyle B$ is 10,000 % of $\displaystyle C$, so $\displaystyle B = \frac{10,000}{100} \cdot C = 100 C$.

$\displaystyle A$ is $\displaystyle \frac{1}{100} \%$, or $\displaystyle 0.01 \%$, of $\displaystyle B$, so $\displaystyle A = \frac{0.01}{100} \cdot B = 0.0001 B$.

Therefore, 

$\displaystyle A = 0.0001 B = 0.0001 \cdot 100 C = 0.01 C$

There are nine "A" tiles, twelve "E" tiles, nine "I" tiles, eight "O" tiles, and four "U" tiles. This is a total of 

$\displaystyle 9 + 12 + 9 + 8 + 4 = 42$ vowel tiles.

12 of the tiles are "E's"; they therefore comprise

$\displaystyle \frac{12}{42} \times 100 \% = \frac{1,200}{42} \% \approx 28.57 \%$.

This rounds to 29%.

13 of the 52 cards in a standard deck are spades. If two spades are removed, then there will remain 11 spades out of 50 cards, or

$\displaystyle \frac{11}{50} \times 100 \% = \frac{1,100}{50} \% = 22 \%$ 

of the remaining cards.

Apply the distributive and commutative properties to the expression in (a):

$\displaystyle -4 (5 - y)$

$\displaystyle = -4 \cdot 5 - \left (-4 \right ) \cdot y$

$\displaystyle = -20 - \left (-4 y\right )$

$\displaystyle = -20+4 y$

$\displaystyle = 4 y +(-20)$

$\displaystyle = 4 y -20$

The two expressions are equivalent.

Apply the distributive property to the expression in (a):

$\displaystyle -5(x+6) = -5 \cdot x+ (-5 ) \cdot 6 = -5x -30$

$\displaystyle -30 < 30$, so $\displaystyle 5x -30 < 5x+ 30$ regardless of $\displaystyle x$.

Therefore, $\displaystyle -5(x+6) < 5x+ 30$

Apply the distributive property to the expression in (a):

$\displaystyle 8(y-5) = 8 \cdot y - 8 \cdot 5= 8y - 40$

Since $\displaystyle -40 < -5$, $\displaystyle 8y - 40 < 8y - 5$, and therefore, regardless of $\displaystyle y$, 

$\displaystyle 8\left ( y - 5 \right )< 8y - 5$

We show that there is at least one value of $\displaystyle y$ that makes the (a) greater and at least one that makes (b) greater:

Case 1: $\displaystyle y = 1$

(a) $\displaystyle -7 (-10+y) = -7 (-10+1) = -7 (-9) = 63$

(b) $\displaystyle 7y + 70 = 7 \cdot 1 + 70 = 7 + 70 = 77$

(b) is greater here

Case 2: $\displaystyle y =- 1$

(a) $\displaystyle -7 (-10-y) = -7 (-10-1) = -7 (-11) = 77$

(b) $\displaystyle 7y + 70 = 7 \cdot \left ( -1 \right )+ 70 = -7 + 70 = 63$

(a) is greater here

We can best solve this by factoring 4 from both terms, and distributing it out:

$\displaystyle 8y + 4z$

$\displaystyle = 4 \cdot 2y + 4 \cdot z$

$\displaystyle = 4 (2y+z)$

$\displaystyle 2 (x+y) +5x$

$\displaystyle = 2 x+2 y +5x$

$\displaystyle = 2 x+5x +2 y$

$\displaystyle =\left ( 2 +5 \right ) x +2 y$

$\displaystyle =7 x +2 y$

(A) and (B) are equivalent variable expressions and are therefore equal regardless of the values of $\displaystyle x$ and $\displaystyle y$.

In order to simiplify we must first distribute the -2 only to what is inside the ( ): 

$\displaystyle -2(x-4)+3x=-2x+8+3x$

Now, we must combine like terms: 

$\displaystyle -2x+8+3x =-2x+3x+8 =x+8$

This gives us the final answer:

$\displaystyle x+8$