\(\displaystyle \dpi{100} \small 2x-3> 0\)

Add 3 to both sides: \(\displaystyle \dpi{100} \small 2x> 3\)

Divide both sides by 2: \(\displaystyle \dpi{100} \small x> \frac{3}{2}\)

To calculate the number of students taking at least 1 science course, add the number of students who are taking each course and subtract the number of students who are taking 2 (to ensure they're not counting twice).

\(\displaystyle \small 80\ +\ 40\ +\ 60\ -\ 13\ =167\)

To calculate the number of students NOT taking a class, subtract this number by the total number of students.

\(\displaystyle \small 200\ -\ 167\ =\ 33\)

\(\displaystyle \small \sqrt[3]{4} \cdot \sqrt{2} = 4^{\frac{1}{3}} \cdot 2^{\frac{1}{2}} = (2^{2})^{\frac{1}{3}} \cdot 2^{\frac{1}{2}} = 2^{2\; \cdot \frac{1}{3}} \cdot 2^{\frac{1}{2}}= 2^{ \frac{2}{3}} \cdot 2^{\frac{1}{2}}= 2^{ \frac{2}{3} + \frac{1}{2}}\)

\(\displaystyle \small = 2^{ \frac{7}{6} }} = \sqrt[6]{2^7} = 2\sqrt[6]{2}\) 

\(\displaystyle \small \frac{1}{x}-\frac{1}{y} = \frac{y-x}{xy}\)

To find \(\displaystyle \small y - x\):

 \(\displaystyle \small \small \small y + x = 9\),

so \(\displaystyle \small \small \left (y + x \right )^{2} =y^{2} + 2xy +x^{2} = 81\)

\(\displaystyle \small \small y^{2} + 2xy +x^{2} - 4xy= 81 - 4 \cdot 16\)

\(\displaystyle \small \small y^{2} - 2xy +x^{2} = (y - x)^{2}= 17\)

\(\displaystyle \small \small y - x = \pm \sqrt{17}\)

Since \(\displaystyle \small \small x < y\), \(\displaystyle \small \small \small \small y - x > 0\)

and we choose the positive square root

\(\displaystyle \small \small \small y - x = \sqrt{17}\)

\(\displaystyle \small \small \small \frac{1}{x}-\frac{1}{y} = \frac{y-x}{xy}= \frac{\sqrt{17}}{16}\)

Today, \(\displaystyle \small C=\frac{B}{3}\). In 3 years, \(\displaystyle \small C=\frac{B}{3}+3\).

The lowest possible value of \(\displaystyle \frac{A}{B}\) is the lowest possible value of \(\displaystyle A\) divided by the highest possible value of \(\displaystyle B\): 

\(\displaystyle \frac{A}{B} \geq 6 \div 6 = 1\)

The highest possible value of \(\displaystyle \frac{A}{B}\) is the highest possible value of \(\displaystyle A\) divided by the lowest possible value of \(\displaystyle B\): 

\(\displaystyle \frac{A}{B} \leq 12 \div 4 = 3\)

\(\displaystyle x^2-y^2 = (x-y)(x+y)\) so this is a composite number for all \(\displaystyle x\) and \(\displaystyle y\).

\(\displaystyle x\cdot y\) is by definition a composite number.

\(\displaystyle x^2y+2y = y\cdot (x^2+2)\) the product of 2 numbers.

This leaves just \(\displaystyle x^2+y^2\).  For a number to be prime, it must be odd (except for 2) so we need to have either \(\displaystyle x\) or \(\displaystyle y\) be odd (but not both).  The first composite odd number is 9.  \(\displaystyle 9^2 = 81\).  The smallest composite number is 4. \(\displaystyle 4^2 = 16\).  

\(\displaystyle 81+16 = 97\) is a prime number.  

So the answer is \(\displaystyle x^2+y^2\)

The definition of the set of real numbers is the set of all numbers that can fit into a/b where a and b are both integers and b does not equal 0. 

So, since we see a fraction here, we know a non-real number occurs if the denominator is 0. Therefore we can find where the denominator is 0 by setting x-3 =0 and solving for x. In this case, x=3 would create a non-real number. Thus our answer is that x CANNOT be 3 for our expression to be a real number. 

This question does not require any calculation. Given that 32 (an even number) is a factor of \(\displaystyle x\), then 2 must be a prime factor. If \(\displaystyle x\) is then multiplied by 2 (to get \(\displaystyle g\)) then \(\displaystyle g\) has no additional unique prime factors (its only additional prime factor, 2, is NOT unique).

\(\displaystyle \dpi{100} \small x+y\) is negative and \(\displaystyle \dpi{100} \small xy\) is positive

\(\displaystyle \dpi{100} \small \frac{Negative}{-Positive} = \frac{Negative}{Negative} = Positive\)

Therefore, the solution cannot be negative.