\(\displaystyle 5xp + 3q = W\)

\(\displaystyle 5xp + 3q -3q = W-3q\)

\(\displaystyle 5xp = W-3q\)

\(\displaystyle 5xp \div 5x =\left ( W-3q \right ) \div 5x\)

\(\displaystyle p =\frac{ W-3q }{5x}\)

\(\displaystyle 5xp + 3q = W\)

\(\displaystyle 5xp + 3q - 3q = W - 3q\)

\(\displaystyle 5xp = W - 3q\)

\(\displaystyle 5xp \div 5p= \left ( W - 3q \right )\div 5p\)

\(\displaystyle x= \frac{W - 3q}{5p}\)

\(\displaystyle 5xp + 3q = W\)

\(\displaystyle 5xp + 3q - 5xp = W- 5xp\)

\(\displaystyle 3q = W- 5xp\)

\(\displaystyle 3q \div 3 = \left ( W- 5xp \right ) \div 3\)

\(\displaystyle q =\frac{W- 5xp }{3 }\)

\(\displaystyle 3x+5x-70=38-4x\)

The first step is to rewrite the problem by combining the like terms.

\(\displaystyle 3x+5x=8x\)

\(\displaystyle 8x-70=38-4x\)

Next we add \(\displaystyle 4x\) to both sides.

\(\displaystyle (8x+4x)-70=38+(-4x+4x)\)

\(\displaystyle 12x-70=38\)

Then we add \(\displaystyle 70\) to both sides to isolate the term with the variable.

\(\displaystyle 12x+ (-70+70)= (38+70)\)

\(\displaystyle 12x=108\)

Finally, we divide both sides by \(\displaystyle 12\).

\(\displaystyle \frac{12x}{12}=x\)

\(\displaystyle \frac{108}{12}=9\)

\(\displaystyle x=9\)

\(\displaystyle 7x^{2}-3=3x^{2}+61\)

The first step is to combine like terms. In this case, we need to subtract \(\displaystyle 3x^{2}\) from both sides.

\(\displaystyle (7x^{2}-3x^{2})-3=(3x^{2}-3x^{2})+61\)

\(\displaystyle 4x^{2}-3=61\)

Next we add \(\displaystyle 3\) to both sides:

\(\displaystyle 4x^{2}+(-3+3)=(61+3)\)

\(\displaystyle 4x^{2}=64\)

Divide both sides by \(\displaystyle 4\):

\(\displaystyle \frac{4x^{2}}{4}=\frac{64}{4}\)

\(\displaystyle \frac{4x^{2}}{4}=x^{2}\)

\(\displaystyle \frac{64}{4}=16\)

\(\displaystyle x^{2}=16\)

Finally, take the square root of both sides:

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

\(\displaystyle \sqrt{16}=4\)

\(\displaystyle x=4\)

With this problem, you'll need to test each of the possible answers to see if they are equivalent to \(\displaystyle \frac{1}{3}\). 

The value of the fraction, if you try and divide it out, has a repeating three after the decimal - which is shown as \(\displaystyle 0.\bar3\). Therefore you can eliminate this answer choice, since it is the same value.

You can also eliminate \(\displaystyle \frac{15}{45}\) because when you reduce that fraction, by dividing the numerator and denominator by 15, you will end up with \(\displaystyle \frac{1}{3}\).

You can also eliminate \(\displaystyle \frac{0.8}{2.4}\). If you multiply the numerator and denominator by a special value of 1 to get rid of the decimal points - in this case multiplying each by 10 - you will get a fraction of \(\displaystyle \frac{8}{24}\). This can more easily be reduced to the fraction in question of \(\displaystyle \frac{1}{3}\).

This leaves us with the correct answer - the value that is not equal to the fraction in question: \(\displaystyle 0.3333333\). While this value is extremely close to the repeating decimal, it is not exactly the same because the fraction does terminate - it has a final ending point. Therefore, it is not the exact same value as \(\displaystyle \frac{1}{3}\).

First, you need to idenitfy what an integer is: any whole number. That can be any positive number, negative number, or zero. A non-integer is a number that cannot be written as a whole number, such as a fraction or decimal number.

You can simplify each expression to determine which one does not simplify into an integer. Three expressions result in integers, while one does not:

\(\displaystyle \sqrt{36} - \sqrt{100} = 6 - 10 = -4\)

\(\displaystyle \sqrt{100-36} = \sqrt{64} = 8\)

\(\displaystyle \sqrt{36} \times \sqrt{100} = 6 \times 10 = 60\)

Each of those three expressions does simplify into an integer, so they are not the correct answers. The following results in a fraction that cannot be simplified into a whole number, so it is not an integer, and thus, the correct answer:

\(\displaystyle \sqrt{100} \div \sqrt{36} = 10 \div 6 = \frac{10}{6} = \frac{5}{3} = 1.\bar{6}\)

To find \(\displaystyle \frac{1}{15}\) of one minute in terms of seconds, you need to first convert one minute to its value in seconds, which is 60 seconds.

Now, you just need to find the value for \(\displaystyle \frac{1}{15}\) of 60, which you determine by multiplying the two values:

\(\displaystyle \frac{1}{15} \times 60 = \frac{60}{15}\)

You can simplify the above fraction by division, which will you give you your correct answer in seconds:

\(\displaystyle \frac{60}{15} = 4\) 

Your answer is 4 seconds.

When adding fractions, generally you determine the least common multiple of the two denominators given to modify the fractions and successfully add. However, this problem gives you a specific denominator to reach: 24.

You will need to change both \(\displaystyle \frac{2}{3}\) and \(\displaystyle \frac{1}{4}\) to fractions that have a denominator of 24. This will require you to modify their numerators to keep the fractions the same. You will then add the two fractions, and the numerator in the sum is your answer.

\(\displaystyle \frac{2}{3} + \frac{1}{4} = \frac{D}{24}\)

First, change \(\displaystyle \frac{2}{3}\) to a fraction of equal value with a denominator of 24. To go from 3 to 24 in the denominator, you must mutliply by 8, so you perform the same with the numerator to keep the fraction at the same value. (You are multiplying by a fraction that is equal to 1, so you do not change the value of the fraction.):

\(\displaystyle \frac{2}{3} \times \frac{8}{8} = \frac{16}{24}\) 

Next, you'll do the same thing with \(\displaystyle \frac{1}{4}\). To go from 4 to 24 in the denominator, you must multiply by 6, so you perform the same operation with the numerator, as described above:

\(\displaystyle \frac{1}{4} \times \frac{6}{6} = \frac{6}{24}\)

Now that you have two fractions with common denominators, you can add them together:

\(\displaystyle \frac{16}{24} + \frac{6}{24} = \frac{D}{24}\)

Adding 16 and 6 together gives you the value of \(\displaystyle D\), which is 22:

\(\displaystyle \frac{22}{24} = \frac{D}{24}\)

You can think of the total number of students in the classroom as an unknown variable, \(\displaystyle x\).

You can set up an equation that includes \(\displaystyle x\) and the information given. Normally in a percent problem, you would start with the total amount of something and multiply it by the percent to get the part of that total amount. You can set this up the same way, but the total amount is the missing information:

Total Amount x Percent = Part of Total Amount

\(\displaystyle x \times\) \(\displaystyle .15\) \(\displaystyle =9\)

You use \(\displaystyle .15\) to numerically represent 15 percent. 

Now you can algebraically solve for \(\displaystyle x\), which will give you the total number of students in the class. You do this by dividing each side by \(\displaystyle .15\).

\(\displaystyle x = 60\)

Therefore there are 60 total students.