Fred has $100 in quarters and nickels initially.  We are also told that he has 260 quarters.  This is worth $65.  Thus Fred initially has $35 in nickels or 700 nickels.

Fred now exchanges some of his nickels for the dimes of a friend.  He ends up with 650 coins.  We know that Fred started with 960 coins (700 nickels + 260 quarters).  He ends up with 650 coins.  The number of quarters remains unchanged, meaning he now has 390 nickels and dimes.  These must have the same value as the initial 700 nickels, though, since he didn't lose any money.

Now we can finally set up our solution:

\(\displaystyle 700 * 0.05= 35 = x*0.05 + (390-x)*0.1 = .05x +39 -.1x = 39 - .05x\)

\(\displaystyle 35 = 39-.05x \Rightarrow .05x = 4 \Rightarrow x = 80\)

Thus Fred has 80 nickels and 310 dimes.

\(\displaystyle \line(1,0){250}\)

An alternative solution step is to notice that turning nickels into dimes always occurs in exactly one way: 2 nickels to 1 dime.  Every time you do this conversion, you will lose exactly one coin.  We then notice that the number of coins drops from 960 to 650, or drops by 310 coins.  We thus need to get rid of 310 coins.  Since we're only allowed to change nickels into dimes (and lose 1 coin each time), we simply do this 310 times to reach the requisite number of coin losses.  We are left with the proper number of coins with the proper value immediately.  Since every replacement replaced 2 nickels, we also lost \(\displaystyle 310\cdot 2=620\) nickels.  Our final number of nickels is thus \(\displaystyle 700-620=80\) nickels.

\(\displaystyle f(-4)=|(-4)^2-20|=|16-20|=|-4|=4\)

Nails are sold in packs of 40 for $1.30.

Screws are $0.07 each.

Since the total is $14.80, we must have a 0 in the hundredths place.  This means that the number of screws must be a multiple of 10 (otherwise it won't be a 0).  Now we are essentially dealing with screws in packs of 10 for $0.70.

\(\displaystyle .70x + 1.30y = 14.80\Rightarrow 14.8-1.3y =.7x\Rightarrow x=\frac{14.8-1.3y}{.7}=\frac{148-13y}{7}\)

Now we have 5 values of \(\displaystyle y\) to check.  If \(\displaystyle 148-13y\) is divisible by 7, then we have the right answer.

What we find is that for \(\displaystyle y=6: \ \ 148-13(6)=70\) is divisible by 7.

Thus our answer is 6 boxes or 240 nails.

The "is" in the question means "equal," so whatever comes before "is" must be equal to whatever comes after. We will find an expression for the information before "is" and an expression for the information after "is," and then we will set these two expressions equal.

Twenty percent of a number can be represented as 0.2n, because 20% expressed as a decimal is 0.2, and because twenty percent "of" a number means the product of that number and twenty percent.

Four greater than the product of a number and six means that we must first find the product of that number and six, and then increase this value by 4.

The product of a number and six means that we must multiply this number by six, which can be represented by 6n. Increasing 6n by 4 can be modeled by the expression 6n + 4, or 4 + 6n (because of the commutative property of addition).

Setting the two expressions equal gives us 0.2n = 4 + 6n .

The easiest way to solve this problem is to guess-and-check the answer choices. The equation that can be used to match the table will be correct.

We can see that the values in the table match the equation \(\displaystyle y=-2x+5\) for each given value. Thus, this must be our answer.

We can also determine certain characteristics from the table itself. For example, as x increases, y(x) decreases. This tells us that there is likely a negative coefficient, which can help narrow down the answer options.

There are B+G total students in the elementary school class, so G out of B+G are girls.

Each term equals the previous term multiplied by \(\displaystyle \frac{3}{2}\).

The fifth term in the sequence is \(\displaystyle \frac{9}{8} \cdot \frac{3}{2} = \frac{27}{16}\).

The sixth term in the sequence is thus \(\displaystyle \frac{27}{16} \cdot \frac{3}{2} = \frac{81}{32}\).

Begin by rearranging the equation to solve for z:

\(\displaystyle \dpi{100} \small z=n+3\)

This means that \(\displaystyle \dpi{100} \small 2z=2\left ( n+3 \right )\), which can be rewritten as \(\displaystyle \dpi{100} \small 2n+6\).

Symbols may appear scary, but these problems are often easier than they first appear. Begin by taking the numbers you are given and replacing the letters in the equation with these numbers.

 \(\displaystyle c\square d = (2c)^{d}\)

\(\displaystyle 4\square 3 = (2(4))^{3}\)

We can now solve the equation. Be sure to perform the operations within the parentheses (PEMDAS) first.

\(\displaystyle 4\square 3=(2(4))^{^{3}}\)

\(\displaystyle 4\square 3=(8)^{3}\)

\(\displaystyle 4\square3 = 512\)

The correct answer is 512.

Begin by solving for \(\displaystyle x\): 

\(\displaystyle \sqrt{3+x}=3\)

First, square both sides of the equation:

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

\(\displaystyle 3+x=9\)

Now, solve for \(\displaystyle x\):

\(\displaystyle 3+x-3 = 9-3\)

\(\displaystyle x=6\)

Now that we know that \(\displaystyle x\) is equal to 6, substitute 6 into the equation we need to solve:

\(\displaystyle x^2-3x+1\)

\(\displaystyle 6^2-3(6)+1\)

Using the correct order of operations (PEMDAS), we find:

\(\displaystyle 6^2-18+1\)

\(\displaystyle 36-18+1\)

\(\displaystyle 18+1\)

\(\displaystyle 19\)

The answer is 19.