TheXSquaredFactor

Yes, 

$7-(-7)=7+7=14$

Remember, subtracting a negative number is the same as adding a positive number. 

24534 attendees who supported queensland!

First of all, let's figure out the percent that were not NSW supporters. To do this, subtract 71% from 100%:

$100\%-71\%=29\%$

29% of people supported queensland. To find the amount of people who supported queensland, multiple the number of attendees by the percent:

$84600*29\%$

$=\frac{84600}{1}*\frac{29}{100}$      Here, I converted the percent to a fraction.

$=\frac{84600*29}{100}$

$=\frac{2453400}{100}=24534$ attendees who supported queensland!

No! You are ignoring order of operations!

The original expression is:

$95/3+23-5*34$

93 is not divisible by three, so we'll leave it for now, but 5*34 can be simplified:

$\frac{95}{3}+23-5*34$

$\frac{95}{3}+23-170$

Now convert the constants to numbers divided by three:

$\frac{95}{3}+\frac{23*3}{3}-\frac{170*3}{3}$

$\frac{95}{3}+\frac{69}{3}-\frac{510}{3}$

$-\frac{346}{3}=-115.333...$

Ignoring order of operations makes a big difference...

B; The solutions are 5, -2, 3i, and -3i

Let's look at the original equation:

$x^4-3x^3-x^2-27x-90=0$

Our goal is to get the left hand side into parts that are more manageable to work with. I'll try to explain this. We'll use the rational root theorem. This says that if a polynomial equation can be written in the form $a_nx^n+a_{n-1}x^{n-1}+...+a_0$ and if $a_0$ and $a_n$ are integers, then the factor can be found by checking numbers produced by doing $\pm\frac{dividers\hspace{1mm}of\hspace{1mm}a_0}{dividers\hspace{1mm}of\hspace{1mm}a_n}$.

Dividers of $a_0:\hspace{1mm}1,2,3,5,6,9,10,15,18,30,45,90$  

Dividers of $a_n:\hspace{1mm}1$

Therefore, check for one rational solution by doing: $\pm\frac{1,2,3,5,6,9,10,15,18,30,45,90}{1}$

-2 happens to be the first value that satisfies this equation. The opposite of that is 2, so a factor is x+2.

$\frac{x^4-3x^3-x^2-27x-90}{x+2}=x^3-5x^2+9x-45$, so the monstrosity we had before goes to $(x+2)(x^3-5x+9x-45)=0$.

Thankfully, this is the hardest part. Now, we have to factor $(x^3-5x^2+9x-45)$ further. Find another factor by grouping:

$(x^3-5x^2)+(9x-45)$

Factor out an x^2 from the first set of parentheses and 9 from the second set and then factor the resulting common term, which happens to be (x-5):

$(x^3-5x^2)+(9x-45)=x^2(x-5)+9(x-5)=(x^2+9)(x-5)$

After all this effort, we have transformed $x^4-3x^3-x^2-27x-90$ to $(x+2)(x-5)(x^2+9)$.

Okay, now use the zero factor principle. In other words, set each factor to zero:

$x+2=0\hspace{1cm}x-5=0\hspace{1cm}x^2+9=0$ 

Let's focus on the easy ones first:

$x=-2$ and $x=5$. 

Now the harder factor:

$x^2+9=0$

$x^2=-9$

$x=\pm\sqrt{-9}$

$x=\pm3i$

Solution set is 5,2,3i, -3i

You need not graph these equations to solve them (but the directions tell you to do so). In case you are in the situation where you are, this is what you can do:

$x^2+y^2=13$

$y=-x+5$

If I were you, I would solve this by substitution. In other words, since y=-x+5, you can substitute that into the y in the other equation, so you would get. Then, simplify:

$x^2+(-x+5)^2=13$

$x^2+x^2-10x+25=13$ 

$2x^2-10x+12=0$

To make things simpler, factor out a 2 because that is the GCF of the left side and then factor (if possible):

$2(x^2-5x+6)=0$      x^2-5x+6 happens to be factorable

$2(x-3)(x-2)=0$     Set each factor set to 0 and solve.

$x-3=0 \hspace{1cm} x-2=0$

$x=3\hspace{1cm}x=2$

Now, plug each x-value to solve for the y-values. I'll pick the second equation as that one has simpler calculations overall:

$y=-(3)+5\hspace{1cm}y=-(2)+5$

$y=2\hspace{1cm}y=3$

Therefore, your solution set is:

$(2,3)$ and $(3,2)$

Let's do that for the other problems for extra practice.

Let's list the original equations for problem 52:

$(x+1)^2+(y-1)^2=18$

$y=x+8$

Substitute x+8 into the first equation and simplify:

$(x+1)^2+((x+8)-1)^2=18$

$(x+1)^2+(x+7)^2=18$

$x^2+2x+1+x^2+14x+49=18$     Combine like terms

$2x^2+16x+32=0$     Factor out a GCF of 2 to simplify things

$2(x^2+8x+16)=0$ 

Now, I could factor x^2+8x+16, but this trinomial is special: it is a perfect square trinomial. I know this because I can work backwards. The x^2 term and 16 term are both perfect squares. The square root of x^2 equals x, and the square root of 16 equals 4. 4x*2=8x. If all three conditions are met (like it is here), then it is a perfect square trinomial:

$2(x+4)(x+4)=0$

$x+4=0$

$x=-4$

In this case, x only has one answer. This is actually easier than the previous problem because we only have to find one y-value instead of two. Now, plug in the x-value into the second equation:

$y=x+8$ 

$y=-4+8$

$y=4$

Therefore, the solution set is:

$(-4,4)$

YOu can apply this method for the next one, too...

Will this satisfy you?

$7^{1997}\approx4.5791609*10^{1687}$

The domain is any x-value that produces a real y-value. In this function every x-value equals -4, so the domain is all the real numbers. We represent that with this notation:

$Domain:{\{x∈\rm I\!R\}}$

The range is the set of all the valid outputs. In this equation, the only valid output is -4. We represent that like this:

$Range:{\{f(x)∈\rm I\!R|f(x)=-4\}}$

To summarize, x can be any real number, and y=-4.

$C=32\pi{m}$

There is a formula for finding the circumference of a circle. It is:

Let C= circumference of circle

Let r= radius of circle
 

$C=2\pi{r}$

Now that we know this formula, let's plug in the information that we know and simplify from there:

$C=2\pi{(16)}=32\pi{m}$

The directions are to leave the answer in terms so pi,so there is no need to provide a decimal approximation. And of course, include units in your final answer

I guess I should take my solution down then because my solution is wrong...

I feel bad for giving a user incorrect info...

You don't have to graph this to figure out the answer. Just know the relationships (although I prefer graphing than remembering the relationships.

A reflection over the x-axis has the following effect:

$(x,y)--->(x,-y)$

A reflection over the y-axis has the following effect:

$(x,y)--->(-x,y)$

Let's try this with point C at (2,2):

First, you reflect over the x-axis. This arrow notation says that the x-coordinate does not change but the y-coordinate changes to -1*y-coordinate:

$(2,2)--->(2,-2)$,so $C'(2,-2)$

Take C' and apply the effect of reflecting over the y-axis:

$(2,-2)--->(-2,-2)$,so $C''(-2,-2)$

Hopefully, this makes sense. This is simply another method of solving. If you want to stick to graphing, that is OK.