If \(\displaystyle x^{2}+ B x+ 60\) is not prime, it is factorable as 

\(\displaystyle (x+m)(x+n)\)

where \(\displaystyle mn = 60\) and \(\displaystyle m + n = B\).

Therefore, we are looking for a whole number that is not the sum of two factors of 60. The integers that are such a sum are

\(\displaystyle 1+60 = 61\)

\(\displaystyle 2 + 30= 32\)

\(\displaystyle 3+20 = 23\)

\(\displaystyle 4+15=19\)

\(\displaystyle 5 +12 = 17\)

\(\displaystyle 6+10 = 16\)

Of the choices, only 18 is not a sum of factors of 60. It is the correct choice.

To simplify the expression start by simplifying each term.

\(\displaystyle \frac{x^{3}}{x}+2x^{2}-5x+\frac{10x}{2}\)

\(\displaystyle ={x^{3-1}}+2x^{2}-5x+5x\)

From here, combine like terms.

\(\displaystyle =x^{2}+2x^{2}\)

\(\displaystyle =3x^{2}\)

\(\displaystyle 3^{2}-10x^{2}+\frac{7^{9}}{7^{7}}\)

To simplify this expression first simplify each expression.

\(\displaystyle =9-10x^{2}+{7^{9-7}}\)

\(\displaystyle =9-10x^{2}+{7^{2}}\)

\(\displaystyle =9-10x^{2}+49\)

From here combine like terms.

\(\displaystyle =-10x^{2}+58\)

This is a simple problem that we can solve by rewriting the denominator as a product of its prime numbers. Since it is an odd number, we can start with 3.

\(\displaystyle \frac{7^{2}\cdot3^{4}\cdot17}{3213}=\frac{7^{2}\cdot3^{4}\cdot17}{3\cdot3\cdot3\cdot119}\)

3213 has three powers of 3. Let's try to divide the rest, 119, by 7.

\(\displaystyle \frac{7^{2}\cdot3^{4}\cdot17}{3\cdot3\cdot3\cdot119}=\frac{7^{2}\cdot3^{4}\cdot17}{3\cdot3\cdot3\cdot7\cdot17}\)

119 is the product of 17 and 7, and since both of these factors are prime numbers, we are done calculating the factors of \(\displaystyle 3213\). Now, we can start canceling factors shared by the numerator and the denominator. After canceling shared factors, we are left with \(\displaystyle 21\), which is the final answer.

\(\displaystyle \frac{7^{2}\cdot3^{4}\cdot17}{3\cdot3\cdot3\cdot7\cdot17}=\frac{7^{2}\cdot3^{4}\cdot17}{3^3\cdot7\cdot17}=7\cdot3=21\)

Firstly, we must find the value of \(\displaystyle x\) for which \(\displaystyle 2^{6x}=4096\) is true. You can solve this simply by testing out powers of \(\displaystyle 2\):

\(\displaystyle 2^{8}= 256\)

\(\displaystyle 2^{9}= 512\)

\(\displaystyle 2^{10}=1024\)

\(\displaystyle 2^{11}=2048\)

\(\displaystyle 2^{12}=4096\)

4096 is 2 raised to the power of 12. Now we can easily say that \(\displaystyle x=2\). We can plug \(\displaystyle 2\) in for \(\displaystyle x\) in \(\displaystyle (2^{x+2})^{3}\) and solve:

\(\displaystyle (2^{2+2})^3=(2^{4})^{3}=2^{12}=4096\)

\(\displaystyle m\) is a multiple of 7, so at the very least it includes a 7. Since \(\displaystyle a\) and \(\displaystyle b\) are both prime numbers, 7 is either \(\displaystyle a\) or \(\displaystyle b\). To make sure we have 49, the square of 7, into our product, we must take both the squares of \(\displaystyle a\) and \(\displaystyle b\) or \(\displaystyle a^{2}b^{2}\), which is the final answer.

Our first two terms each involve an exponent in parentheses raised to an exponent outside the parentheses, so in this case, for each term, we multiply the exponent inside the parentheses with the exponent outside the parentheses. For the third term, we have the same two numbers raised to a certain power, so we add the exponents. This gives us:

\(\displaystyle (2^2)^2+(3^2)^2+32(\frac{1}{2})^2(\frac{1}{2})^3\)

\(\displaystyle =2^4+3^4+32(\frac{1}{2})^5\)

We can then solve the expression:

\(\displaystyle =16+81+32(\frac{1}{32})\)

\(\displaystyle =16+81+1\)

\(\displaystyle =98\)

Recall the rule dealing with raising exponents to a higher power when an exponent in parenthesis is raised to an exponent outside of the parentheses; multiply the exponents together to get the new exponent. Don't be confused by the fractional exponents. Simply multiply across the numerator and across the denominator.

\(\displaystyle \small \small (z^{\frac{4}{3}})^{\frac{6}{8}}=z^\frac{4*6}{3*8}=z^{\frac{24}{24}}=z^1=z\)

In this case, the expression simplifies down to just \(\displaystyle z\).

For any positive integer whose last digit is \(\displaystyle a\), the last digit if this integer raised to any power \(\displaystyle p\) is the same as the last digit of \(\displaystyle a^{P}\). For our problem, the last digit is then given by \(\displaystyle 7^{43}\). Now, this is pretty complicated to calculate, so we can try to find a pattern in the last digits of the powers of 7.

\(\displaystyle 7^{1}\) is \(\displaystyle 7\)

\(\displaystyle 7^{2}\) is \(\displaystyle 9\)

\(\displaystyle 7^{3}\) is \(\displaystyle 3\)

\(\displaystyle 7^{4}\) is \(\displaystyle 1\)

\(\displaystyle 7^{5}\) is \(\displaystyle 7\) again.

\(\displaystyle 7^6\) is 9 again.

So, the pattern repeats every 4 numbers. Therefore, if we divide the power of 43 in \(\displaystyle 7^{43}\) by 4, we get a remainder of 3. Therefore, the final answer is the last digit of \(\displaystyle 7^{3}\), or \(\displaystyle 3\).

For any positive integer whose last digit is \(\displaystyle a\), the last digit if this integer raised to any power \(\displaystyle p\) is the same as the last digit of \(\displaystyle a^{P}\). So, the last digit of \(\displaystyle 1239^{56}\) is given by its last digit raised to the power of 56, or \(\displaystyle 9^{56}\).

Let's then try to find a pattern in the last digits of the powers of 9.

\(\displaystyle 9^{1}\) is \(\displaystyle 9\)

\(\displaystyle 9^{2}\) is \(\displaystyle 1\)

\(\displaystyle 9^{3}\) is \(\displaystyle 9\)

We can see that the odd powers of 9 have 9 as their last digits and the even powers of 9 have 1 for their last digits.

Since 56 is an even number, the last digit of \(\displaystyle 1239^{56}\) is \(\displaystyle 1\).