First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

\(\displaystyle (-\frac{5}{3})^{-3}=(-\frac{3}{5})^3\)

Apply the exponent within the parentheses and simplify. The negative in the fraction can be applied to either the numerator or the denominator, but not both; we will apply it to the numerator.

\(\displaystyle \frac{(-3)^3}{(5)^3}\)

\(\displaystyle \frac{-27}{125}=-\frac{27}{125}\)

The fraction cannot be simplified further.

\(\displaystyle \dpi{100} \frac{1}{3}x-14=7\)

\(\displaystyle \dpi{100} \frac{1}{3}x-14+14=7+14\)

\(\displaystyle \dpi{100} \frac{1}{3}x=21\)

\(\displaystyle \dpi{100} 3\cdot \frac{1}{3}x=21\cdot 3\)

\(\displaystyle \dpi{100} x=63\)

If the product of three consecutive numbers is 990, then the smallest number can be found by plugging in each answer choice into the scenario to see whether it is correct. 

If we plug in 9 as the smallest number, then the two consecutive numbers would be 10 and 11. 

Given that 9 times 10 times 11 equals 990, that is the correct answer. 

The first step is to simplify the values in the parentheses:

\(\displaystyle (\frac{5}{4}\cdot8+6\div3)^{2}-12\)

\(\displaystyle (10+2)^{2}-12\)

\(\displaystyle 12^{2}-12\)

\(\displaystyle 144-12\)

\(\displaystyle 132\)

Reorder the expression to group like-terms together.

\(\displaystyle 4xy^2-2xy+y^{2}x+y^3-8xy+4yx\)

\(\displaystyle (4xy^2+y^{2}x)+(-2xy+-8xy+4yx)+y^3\)

\(\displaystyle 5xy^2+(-6xy)+y^3\)

\(\displaystyle 5xy^2-6xy+y^3\)

If x is equal to \(\displaystyle -4\), then the equation could be written as follows:

\(\displaystyle x^{2}-12x\)

\(\displaystyle -4^{2}-(12*-4)\)

\(\displaystyle 16+48=64\)

Given that \(\displaystyle 64\) is a positive number, \(\displaystyle -4\) is a possible value of \(\displaystyle x\). 

In order to solve this problem, the price of $50 must be discounted by 10% until we get to a price of $38 or less. 

The first class is 50. 

The second class, being 10 percent less than the previous one, is \(\displaystyle 50-.1\cdot 50=50-5=45\)

The third class, being 10 percent less than the previous one, is \(\displaystyle 45-.1\cdot45=45-4.5 = 40.5\)

The fourth class, being 10 percent less than the previous one, is \(\displaystyle 40.5-.1\cdot40.5=40.5-4.05=36.45\)

Therefore, the answer is 4. 

Simplify the following expression

\(\displaystyle (6x^2)*(3x^3)\)

Let's begin by multiplying our coefficients:

\(\displaystyle 6*3=18\)

Next, we need to realize that we can combine our x's by adding the exponents.

\(\displaystyle x^2*x^3=x^{3+2}=x^5\)

Put it all together to get:

\(\displaystyle 18x^5\)

Simplify the following:

\(\displaystyle (9q^3n^3)(q^5n^6)\)

To begin, our coefficient will not change. We have just one integer (the 9) and nothing to multiply it by.

To combine our exponents, we will add them. This is because we are multiplying them

\(\displaystyle q^3*q^5=q^{3+5}=q^8\)

\(\displaystyle n^3*n^6=n^{3+6}=n^9\)

Put it together to get:

\(\displaystyle 9q^8n^9\)

Multiply the binomials using the FOIL method - first, outer, inner, last - then combine like terms:

\(\displaystyle (2x+ 3y) (3x-2y)\)

\(\displaystyle = 2x \cdot 3x - 2x \cdot 2y+ 3y \cdot 3x-3y \cdot 2y\)

\(\displaystyle =6 x ^{2}-4 x y+ 9xy-6y^{2}\)

\(\displaystyle =6 x ^{2}+ 5 xy-6y^{2}\)

\(\displaystyle x^{2} = 18\) and \(\displaystyle y^{2} = 2\); also, by the Power of a Product Principle:

\(\displaystyle (xy) ^{2} = x^{2}y^{2}\).

\(\displaystyle x\) and \(\displaystyle y\) are both positive, so, substituting:

\(\displaystyle xy = \sqrt{x^{2}y^{2}} = \sqrt{18 \cdot 2} = \sqrt{36} = 6\).

Again, using substitution:

\(\displaystyle (2x+ 3y) (3x-2y)\)

\(\displaystyle =6 x ^{2}+ 5 xy-6y^{2}\)

\(\displaystyle =6 (18)+ 5 (6)-6 (2)\)

\(\displaystyle =108 +30-12\)

\(\displaystyle =126\)