TheXSquaredFactor

Your calculator is correct; the expressions evaluates to 9.

Be careful with order of operations. Remember, there are strict rules when evaluating expressions. 
 

1. Evaluate Parentheses from left to right

2. Evaluate Exponents from left to right

3. Evaluate Multiplication or Division from left to right

4. Evaluate Addition or Subtraction from left to right

Okay, with these rules in place let's try to evaluate this expression

I think I know how you got your answer of 1:

Just be vigilant on how you evaluate expressions because a mistake like the one I illustrated above can give you a wildly different answer. 

\(\frac{16}{\frac{1}{4}}=64\). In factor form, \(64=2^6\)

\(\frac{16}{\frac{1}{4}}\)

This is the original expression given, or, at least, this was my interpretation. If it is wrong, please correct me. Dividing a number by a fraction is the same as multiplying a number by that fraction's reciprocal:

\(\frac{16}{\frac{1}{4}}=16*\frac{4}{1}=64\)

You said that you wanted your answer in factor form. Continue creating a tree of factors until you get down to prime numbers:

64

/\

8   *   8

/\         /\

2 * 4    4 * 2

               /\     /\                

2*2   2*2

Combine the bolded numbers, and you are done! There are 6 2's. 

\(2*2*2*2*2*2=2^6\)

No, \(1^{127}=1\). 1 is not a prime number. 

Why? To understand, let's make a definition for a prime number. A prime number is a whole number that has exactly two unique positive whole-numbered divisors.

Does 1 fit into this definition? No, it does not. To get one, you can multiply 1 and 1 together. 1 and 1 are not unique divisors. Therefore, 1 is not a prime number.

Ignore this message...

Number 1:5

Number 2:8

In this problem, "one number" and "another number" are simply referring to variables like x and y. "Is" means equal to. Therefore, twice one number is simply 2 multiplied by a variable. For my variables, I'll simply use x and y.

\(2x+y=18\)

\(4x-y=12\)

This is a system of equations. Normally, I gravitate toward substitution, but I'll actually use elimination to solve this. Notice how there is a y-term and a negative y-term. I can add the equations together to get this:

\(6x=30\)

Divide by 6 on both sides:

\(x=5\)

Okay, our first number is equal to five. To find the next number, y, substitute 5 into one of the equations. It doesn't matter which one; you'll get the same answer:

\(2(5)+y=18\)

Simplify 2*5:

\(10+y=18\)

Subtract 10 on both sides to isolate y, our second number:

\(y=8\)

Ah ha! I got it after trying for some time... It's 78!

Flip your monitor upside down. What pattern do you see now? When you orient the numbers in the correct position, this is what you see...

\(86, ?,88,89,90,91\)

The numbers are going in ascending by one each time. The ? is clearly 87. Let's flip this number back around to see what it looks like. Flip the order to make the number look like it would in the sequence. You get 78!

smart cookie....

\(19.35*10=193.5\)

When multiplying by 10, the decimal place shifts to the right one place.

As a generalization, \(x^0=1\hspace{1cm},x\neq0\)

This table may help you understand why 3^0=1

Do you notice a pattern? I do. As you go down the list, you can divide by three to get to the next value. Therefore, if 3^1=3, all you have to do to get the next value is to divide by three. 3^1/3=1, so 3^0=1.

Here's another way of thinking about it. This method works for any number to the power of 0:

\(1=\frac{x^n}{x^n}=x^{n-n}=x^0\hspace{1cm},x\neq0\)

This is the quadratic formula:
 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Let's try it with an equation to see it in action:

\(x^2+6x-3=0\)

First, find a,b, and c and take note of those values! Remember, a typical quadratic can be written in the form of

\(ax^2+bx+c=0\). a equals the coefficient of the x^2 term., b equals the coefficient of the x-term, and c equals the constant.

a=1, b=6, c=-3

Now, plug those values into the quadratic formula and solve for x:

\(x = {-6\pm \sqrt{6^2-4(1)(-3)} \over 2(1)}\)

\(x = {-6 \pm \sqrt{36+12)} \over 2}\)

\(x = {-6 \pm \sqrt{48} \over 2}\)

\(x = {-6\pm 4\sqrt{3} \over 2}\)

\(x = -3\pm 2\sqrt{3}\)

Now, try this example on your own:

\(-3x^2-x+14=0\)