TheXSquaredFactor

Well, thank you!

Sure, I think I can give you a fairly simple proof for you. There are a few property of exponents that you have learned. I have listed all of them for you and what each says. To understand these proofs, you should understand that for whole numbers of a, \(x^a=\underbrace{x*x*x*...*x}\\ \quad\quad\quad\quad\text{a times}\). This is the basis of the notation. The proofs are based on the notation. Sometimes, some rules will assume that a previous one was already proven.

I will write the proofs in order.

1) \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=\underbrace{\underbrace{x*x*x*...*x}*\underbrace{x*x*x..*x}\\ \hspace{7mm}\text{a times}\hspace{18mm}\text{b times}}\\ \hspace{40mm}\text{a+b times}\\ \), so \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}+{\textcolor{blue}{b}}}\)

2) \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=\frac{\underbrace{x*x*x*...*x}\\ \hspace{2mm}\text{a times}}{\underbrace{x*x*x*...*x}\\ \hspace{2mm}\text{b times}}\). Here, the x in the numerator will cancel out the x in the denominator a-b times, so \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=x^{\textcolor{red}{a}-\textcolor{blue}{b}}\)

3a) 

3b) 

4) \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=\underbrace{x^{\textcolor{red}{a}}*x^{\textcolor{red}{a}}*x^{\textcolor{red}{a}}*...*x^{\textcolor{red}{a}}}= x^{\underbrace{{\textcolor{red}{a}}+{\textcolor{red}{a}}+{\textcolor{red}{a}}+...+{\textcolor{red}{a}}}}\\ \hspace{27mm}\text{b times}\hspace{23mm}\text{b times}\) 

If there are b lots of a, this is equivalent to multiplication, so \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}\textcolor{blue}{b}}\)

5) \((xy)^\textcolor{red}{a}=\underbrace{xy*xy*xy*...*xy}=\underbrace{(x*x*x*...*x)}*\underbrace{(y*y*y...*y)}=x^\textcolor{red}{a}y^\textcolor{red}{a}\\ \hspace{27mm}\text{a times}\hspace{27mm}\text{a times}\hspace{22mm}\text{a times}\)

If we multiply lots of x and y a times, then we can convert this back to an exponent.

6a) \(\left(\frac{x}{y}\right)^\textcolor{red}{a}=\underbrace{\frac{x}{y}*\frac{x}{y}*\frac{x}{y}*...*\frac{x}{y}} =\underbrace{(x*x*x*...*x)}*\underbrace{\left(\frac{1}{y}*\frac{1}{y}*\frac{1}{y}*...*\frac{1}{y}\right)} =\frac{x^\textcolor{red}{a}}{1}*\frac{1}{y^\textcolor{red}{a}}=\frac{x^\textcolor{red}{a}}{y^\textcolor{red}{a}}\\ \hspace{27mm}\text{a times}\hspace{22mm}\text{a times}\hspace{25mm}\text{a times}\)

6b) 

 Hopefully, these proofs will aide you in your math journey. Always question why something is and don't accept them as fact!

\(\frac{x+6}{x-3}=\frac{2x}{x-3}\)

Since the denominators are already the same, all you need to do is set the numerators equal to each other. 

\(x+6=2x\\ 6=x\)

This method is significantly more elegant than my long-winded answer. 

\(f(x)=ax^2+bx+a\) 

If one root is located at x=4, then we can find an equation for this.

Since we now know what b equals in terms of a, we can simplify the original function to \(f(x)=ax^2-\frac{17a}{4}x+a\). We can now begin to solve this and find the other root.

Therefore, x=1/4 is the other zero of this polynomial function.

Let's break this equation down step-by-step. Let's start with the solid line that goes upward from left to right first. The most useful type of linear equation is slope-intercept because it is easy to build the equation based on individual features of each linear equation.Let's pretend, for now, that the lines are not inequalities and instead equations.

This immediately eliminates the last option since the inequality symbol is inverted, which in incorrect. Unfortunately, though, we can manipulate this answer and get the rest of the answers, so we will have to move on to the next line. Let's do the solid line that goes downward from left to right. Let's utilize the exact same process as before. 

We can eliminate the first answer choice once again because the sign is incorrect. We can also eliminate the second option choice, too, because it is incorrect as well. It is 

\(-y\leq 2x+8\)

This is invalid because when multiplying or dividing by a negative number, the inequality sign is flipped, so this is incorrect, too. 

Therefore, there is no need to check for the other line, so the third option is correct.

I found my error. It was a dumb one, as always.

Thanks! 

We already know the values of the given variables, so simply substitute the known value for the given value into the expression and evaluate carefully.

This one may not be nearly as obvious as the previous problem. The given sequence is definitely an arithmetic sequence because there is a common difference. The general formula for an arithmetic sequence is the following:
 

\(a_n=d(n-1)+c\)

d = the common difference

c = the first term in the sequence (sometimes referred to as a₁)

The common difference is the constant amount of change between numbers. In this case, there exists a common differnece.

\(\underbrace{9, 13}\underbrace{, 17},...\\ +4\hspace{3mm}+4\)

Since the common diffence is 4, let's put that in for d.

\(a_n=4(n-1)+c\)

The first term in this sequence is given; it is 9.

\(a_n=4(n-1)+9\)

Just like that, we have generated the formula that generates the nth term in the sequence. To figure out the 48th term, simply evaluate at n=48.

\(a_{48}=4(48-1)+9\\ a_{48}=4*47+9\\ a_{48}=197\)

The sequence above follows a pattern.

\(f(n)=7n\)

f(n) is the output of the nth term in the sequence

n is the nth term in the sequence.

Since we have established this pattern, we can use this to figure out the 42nd term. If we know that 7n represents the value of the sequence at the nth term in the sequence, then we can use it to find the 42nd term.

\(7*42=294\)

294 is the 42nd term in the sequence.