$\displaystyle \frac{x^{3}-2x^{2}-3x+5}{5x+3}$

1. Plug in $\displaystyle 5$ wherever there is an $\displaystyle x$ in the above equation.

$\displaystyle \frac{5^{3}-(2)(5^{2})-(3)(5)+5}{(5)(5)+3}$

2. Perform the above operations.

$\displaystyle \frac{5^{3}-(2)(25)-(3)(5)+5}{(5)(5)+3}$

$\displaystyle \frac{125-50-15+5}{25+3}$

$\displaystyle \frac{65}{28}=2.32$

Since Mary traveled 3 times as quickly coming from school as she did going to school (6 miles per hour compared to 2 miles per hour), we know that Mary spent only a third of the time coming from school as she did going. If x represents the number of hours it took to get to school, then x/3 represents the number of hours it took her to return.

Knowing that the total trip took 1 hour, we have:

x + x/3 = 1

3x/3 + 1x/3 = 1

4x/3 = 1

x  = 3/4

So we know it took Mary 3/4 of an hour to travel to school (and the remaining 1/4 of an hour to get back).

Remembering that distance =  rate * time, the distance Mary traveled on her way to school was (2 miles per hour) * (3/4 of an hour) = 3/2 miles. Furthermore, since she took the same route coming back, she must have traveled 3/2 of a mile to return as well.

Therefore, the the total number of miles in Mary's round trip is 3/2 miles + 3/2 miles = 6/2 miles = 3 miles.

To raise $\displaystyle \frac{1}{8}$ to the exponent $\displaystyle \frac{2}{3}$, square $\displaystyle \frac{1}{8}$ and then take the cube root.

$\displaystyle w^\frac{2}{3}=\sqrt[3]{w^2}$

$\displaystyle (\frac{1}{8})^\frac{2}{3}=\sqrt[3]{(\frac{1}{8})^2}=\sqrt[3]{\frac{1}{64}}$

$\displaystyle \sqrt[3]{\frac{1}{64}}=\frac{1}{4}$

$\displaystyle (\frac{1}{8})^{\frac{2}{3}}=\frac{1}{4}$

The common denominator of the left side is x(x–1). Multiplying the top and bottom of 1/x by (x–1) yields

Since this statement is true, there are infinitely many solutions. 

To begin a problem like this, you must first convert everything to $\displaystyle \sin x$ and $\displaystyle \cos x$ alone. This way, you can begin to cancel and combine to its most simplified form.

Since $\displaystyle \csc x=\frac{1}{\sin x}$ and $\displaystyle \cot x=\frac{\cos x}{\sin x}$, we insert those identities into the equation as follows.

$\displaystyle \sin ^{2}x\cdot \frac{1}{\sin x}\cdot \frac{\cos x}{\sin x}$

From here we combine the numerator and denominators of each fraction together to easily see what we can combine and cancel.

$\displaystyle \frac{\sin^{2}x\cdot \cos x}{\sin x\cdot \sin x}$

Since there is a $\displaystyle \sin ^{2}x$ in the numerator and the denominator, we can cancel them as they divide to equal 1. All we have left is $\displaystyle \cos x$, the answer. 

Solve for the variables, the plug into formula.

x = 12/3 = 4

y = 10 * 4 = 40

z= 9/4 = 2 ¹/₄

10xyz = 3600

Xy = 160

3600/160 = 22 ¹/2

In order to solve $\displaystyle \frac{2x+y}{z}$, we must first find the values of $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$ using the initial equations provided. Starting with $\displaystyle 5x=55$:

$\displaystyle x=\frac{55}{5}=11$

Then:

 $\displaystyle x+y=23$

$\displaystyle 11+y=23$

$\displaystyle y=12$

Finally:

$\displaystyle y-z=2$

$\displaystyle 12-z=2$

$\displaystyle 10=z$

With the values of $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$ in hand, we can solve the final equation:

$\displaystyle \frac{2x+y}{z}$

$\displaystyle =\frac{2(11)+12}{10}$

$\displaystyle =\frac{22+12}{10}$

$\displaystyle =\frac{34}{10}$

$\displaystyle =\frac{17}{5}$

In order to solve $\displaystyle \frac{2}{x+y}$, first substitute the values of $\displaystyle x$ and $\displaystyle y$ provided in the problem:

$\displaystyle \frac{2}{\frac{1}{2}+\frac{2}{3}}$

Find the Least Common Multiple (LCM) of the fractional terms in the denominator and find the equivalent fractions with the same common denominator:

$\displaystyle \frac{2}{\frac{1}{2}+\frac{2}{3}}$

$\displaystyle =\frac{2}{\frac{3}{6}+\frac{4}{6}}$

$\displaystyle =\frac{2}{\frac{7}{6}}$

Finally, in order to divide by a fraction, we must multiply by the reciprocal of the fraction:

$\displaystyle =2\times \frac{6}{7}$

$\displaystyle =\frac{12}{7}$

In order to solve for $\displaystyle w$, first substitute $\displaystyle y=\frac{1}{9}$ into the equation for $\displaystyle w$:

 $\displaystyle w=\bigg(\frac{1}{3}+y\bigg)^{2}$

$\displaystyle w=\bigg(\frac{1}{3}+\frac{1}{9}\bigg)^{2}$

Then, find the Least Common Multiple (LCM) of the two fractions and generate equivalent fractions with the same denominator:

$\displaystyle w=\bigg(\frac{3}{9}+\frac{1}{9}\bigg)^{2}$

Finally, simplify the equation:

$\displaystyle w=\bigg(\frac{4}{9}\bigg)^{2}$

$\displaystyle w=\frac{16}{81}$

Factor out 7 from the numerator: $\displaystyle \frac{7(7^{11}-7^{9})}{7^{11}-7^{9}}$

This simplifies to 7.