Subtract the two numbers, being careful not to forget to remove the borrowed number.

You can also add the solution back to 1296 to check your work, as addition is easier than subtraction.

$\displaystyle \dpi{100} 1296+3297=4593$

First convert $\displaystyle \dpi{100} \frac{9}{2}$ into a decimal. 

$\displaystyle \dpi{100} 9\div 2=4\ remainder\ of\ 1$

So we are left with $\displaystyle \dpi{100} 4\frac{1}{2}$, which is 4.5 in decimal form.

Now subtract:

$\displaystyle \dpi{100} 4.8-4.5=0.3$

You set up the expression with $\displaystyle \left ( 5a-4b-3c\right )$ on the top and $\displaystyle \left ( 2a-3b+7c\right )$ on the bottom. Because this is subtraction, remember to distribute the negative sign to the expression on the bottom, and then add, so

you get $\displaystyle 3a-b-10c$.

The prime integers between 65 and 75 are 67, 71, and 73, so $\displaystyle K$ assumes one of those values; the prime integers between 45 and 55 are 47 and 53, so $\displaystyle L$ assumes one of those values. Therefore, one of the following holds true:

$\displaystyle K - L = 73 - 53 = 20$

$\displaystyle K - L = 73 - 47 = 26$

$\displaystyle K - L = 71 - 53 = 18$

$\displaystyle K - L = 71 - 47 = 24$

$\displaystyle K - L = 67 - 53 = 14$

$\displaystyle K - L = 67 - 47 = 20$

There are five possible values for $\displaystyle K - L$ (20 appears twice here).

The greatest possible value of $\displaystyle L - K$ is the least possible value of $\displaystyle K$ subtracted from the greatest possible value of $\displaystyle L$. The least prime between 55 and 65 is 59, and the greatest prime between 85 and 95 is 89, so $\displaystyle L = 89$ and $\displaystyle K = 59$ give the greatest possible value of $\displaystyle L - K$, which is equal to $\displaystyle 89-59 = 30$

$\displaystyle a \Upsilon b = a^{5} - b^{4}$

$\displaystyle \frac{1}{2}\; \Upsilon\; \frac{1}{2} =\left ( \frac{1}{2} \right )^{5} - \left ( \frac {1}{2} \right )^{4}$

$\displaystyle = \frac{1}{32} - \frac{1}{16}$

$\displaystyle = \frac{1}{32} - \frac{2}{32}$

$\displaystyle = -\frac{1}{32}$

$\displaystyle g(x)=\left | x^{2} - \frac{1}{x^{2}}\right |$

$\displaystyle g(2)=\left | 2^{2} - \frac{1}{2^{2}}\right | = \left | 4- \frac{1}{4}\right | = \left | 3 \frac{3}{4}\right |= 3 \frac{3}{4}$

$\displaystyle g\left ( \frac{1}{2} \right )=\left | \left ( \frac{1}{2} \right )^{2} - \frac{1}{\left ( \frac{1}{2} \right )^{2}}\right | = \left | \frac{1}{4}- \frac{1}{ \frac{1}{4}} \right | = \left | 3 \frac{3}{4}\right | = \left | \frac{1}{4}- 4 \right | = \left | - 3 \frac{3}{4} \right | = 3 \frac{3}{4}$

$\displaystyle g(2) - g \left ( \frac{1}{2} \right ) = 3 \frac{3}{4} - 3 \frac{3}{4} = 0$

$\displaystyle a \forall b = a^{4} + \frac{1}{b^{4}}$

$\displaystyle \frac{1}{2} \forall (-2 ) = \left (\frac{1}{2} \right )^{4} + \frac{1}{(-2)^{4}}= \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$

$\displaystyle g(x)=\left | x^{5}-x^{2}\right |$

$\displaystyle g(2)=\left | 2^{5}-2^{2}\right | = \left | 32-4\right | = \left | 28\right | = 28$

$\displaystyle g(-2)=\left |( -2)^{5}-\left (-2 \right )^{2}\right | = \left | -32-4\right | = \left | -36\right | = 36$

$\displaystyle g(2) - g(-2) = 28 - 36 = -8$

$\displaystyle f(x) = 10 - \sqrt[3]{x} - \sqrt[5]{x}$

$\displaystyle f(-1) = 10 - \sqrt[3]{-1} - \sqrt[5]{-1} = 10 - (-1)- (-1)= 10+1+1 = 12$