Here is a diagram for you to reference as I solve for the area of the triangle. Notice that I added \(\overline{FG}\) such that the segment is the altitude of the triangle.
Since the altitude and base of this triangle lie on gridlines, we can simply count the distance (as opposed to using the distance formula).
\(ED=6\\ FG=4\)
We already have all the information needed to solve for the area. Just use the area formula for a triangle, A=1/2*bh.
\(A_{\triangle DEF}=\frac{1}{2}ED*FG\) | We already know the values of ED and FG, so let's plug them in! |
\(A_{\triangle DEF}=\frac{1}{2}*6*4\) | Now, we simplify. multiplying 6 by 1/2 eliminate the fraction, but you can multiply in any order you'd like. |
\(A_{\triangle DEF}=3*4\) | |
\(A_{\triangle DEF}=12\text{units}^2\) | |
Let's line the equation up side by side.
\(\hspace{4mm}\textcolor{red}{4x}+8y=+20\\ \textcolor{red}{-4x}+2y=-30\)
Take note of the bit highlighted in red. Do you notice how the coefficients are opposite of each other? This is important! If I were to add these equations together as is, then the x would cancel out.
\(\hspace{4mm}\textcolor{red}{4x}+8y=+20\\ \textcolor{red}{-4x}+2y=-30\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad}\\ \hspace{2mm}0x+10y=-10\)
We can solve for y pretty easily by dividing by -10. Of course, 0x simplifies to 0, so the x has disappeared.
\(10y=-10\\ \hspace{5mm}y=-1\)
We can now plug in this y value into the the original equation and solve for x. I'll plug it into the first one.
\(4x+8y=20\) | We already know that y=-1, so let's substitute that in for y and solve for the remaining unknown. |
\(4x+8(-1)=20\) | 8*-1 can be simplified. |
\(4x-8=20\) | Add 8 to both sides. |
\(4x=28\) | Finally, divide by 4 to isolate x completely. |
\(x=7\) | |
If the roots are \(x=2-\sqrt{3}\) and \(x=1-\sqrt{3}\), then the conjugates must also be roots as well, in order for it to be a polynomial. Therefore, \(x=2+\sqrt{3}\) and \(x=1+\sqrt{3}\) are also roots of this unknown polynomial.
With these roots, we can generate factors of the given polynomial, too.
\(x=2-\sqrt{3}\\ \hspace{1mm}-\left(2-\sqrt{3}\right)\) | \(x=2+\sqrt{3}\\ \hspace{1mm}-\left(2+\sqrt{3}\right)\) | \(x=1-\sqrt{3}\\ \hspace{1mm}-\left(1-\sqrt{3}\right)\) | \(x=1+\sqrt{3}\\ \hspace{1mm}-\left(1+\sqrt{3}\right)\) | By subtracting the right hand side of the equation, we can then see what the factors are. |
\(x-\left(2-\sqrt{3}\right)=0\) | \(x-\left(2+\sqrt{3}\right)=0\) | \(x-\left(1-\sqrt{3}\right)=0\) | \(x-\left(1+\sqrt{3}\right)=0\) | |
\(x-2+\sqrt{3}=0\) | \(x-2-\sqrt{3}=0\) | \(x-1+\sqrt{3}=0\) | \(x-1-\sqrt{3}=0\) | Since all of these are set equal to zero, these all must be factors of the original unknown polynomial. |
As aforementioned, they are factors of the original equation. If they are factors, then we can multiply them together and figure out what the original polynomial is.
\(\textcolor{red}{\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)}\textcolor{blue}{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}=0\)
We know that the left hand side is equal to zero since each factor equals zero. 0 multiplied by itself four times still equals zero. Now, expand. I recommend multiplying factors that are conjugates. This might make the process easier. We can multiply in any order, too. I'll expand the bit in red first.
\(\textcolor{red}{\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)}\) | Make sure to remain attentive. Be sure to multiply every factor. |
\(\textcolor{red}{x^2-2x-x\sqrt{3}-2x+4+2\sqrt{3}+x\sqrt{3}-2\sqrt{3}-3}\) | Wow! Those are a lot of terms. Let's rearrange the equation, though, to see if any cancelling can occur here. (Spoiler: There is!) |
\(\textcolor{red}{x^2-x\sqrt{3}+x\sqrt{3}+2\sqrt{3}-2\sqrt{3}-2x-2x+4-3}\) | Look at that! All the radical expression cancel out! |
\(\textcolor{red}{x^2-4x+1}\) | Wow, that monstrosity has been simplified quite beautifully. Don't you agree? |
Let's do the exact same process with the blue bit. It should also simplify to a quadratic trinomial.
\(\textcolor{blue}{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}\) | Ok, time to expand again! We already have an idea of what should occur. |
\(\textcolor{blue}{x^2-x-x\sqrt{3}-x+1+\sqrt{3}+x\sqrt{3}-\sqrt{3}-3}\) | Let's do that rearranging again! |
\(\textcolor{blue}{x^2-x-x-x\sqrt{3}+x\sqrt{3}+\sqrt{3}-\sqrt{3}+1-3}\) | Yet again, as expected, the radicals will cancel out. |
\(\textcolor{blue}{x^2-2x-2}\) | |
Now, we must multiply the red and blue parts together to get a quartic polynomial.
\((\textcolor{red}{x^2-4x+1})(\textcolor{blue}{x^2-2x-2})=0\) | We have to expand one more time! |
\(x^4-2x^3-2x^2-4x^3+8x^2+8x+x^2-2x-2=0\) | Yet again, let's rearrange and see if any combining can take place. |
\(x^4-2x^3-4x^3-2x^2+8x^2+x^2+8x-2x-2=0\) | Ok, now let's combine. |
\(x^4-6x^3+7x^2+6x-2=0\) | We can turn this into a function. |
\(f(x)=x^4-6x^3+7x^2+6x-2\)
Let's check to see that this function fits the strict conditions given in the original problem.
This polynomial fits these conditions, so this is the above is the answer.
17)
Instead of using traditional methods such as actually solving the system of equations, you can actually use process of elimination quite easily here to determine the solutions. This is most likely faster than actually solving the system of equations.
There are two answers with the solution set of (-4,0), but this cannot be right since it does not satisfy the first equation. Therefore, the first and last option are already not correct. The numbers are so easy to work with that this is easy to see.
The third option has (0,-4), which does not satisfy the first equation either.
Therefore, the second answer choice must be correct. Wow, this solution involved minimal math.
18)
The zeroes of the function means that y is set equal to 0.
\(0=(x-3)(x+2)(x-2)\)
Since the factors of the original polynomial are already given to us, simply set the factors equal to 0 and solve.
\(x-3=0\) | \(x+2=0\) | \(x-2=0\) |
\(x=3\) | \(x=-2\) | \(x=2\) |
These values for x correspond to third option choice listed.
1)
The first question is an example of a geometric progression since the given sequence has a constant ratio of 4. This can be determined by finding the quotient of any term of the sequence and the immediately preceeding term. The general structure of geometric progressions are of the following:
\(a_n=a_1*r^{n-1}\)
n = the nth term of the given geometric progression
r = the common ratio
a1= the first term of the sequence
Knowing all of this, we can substitute a few values into this formula.
\(a_n=3*4^{n-1}\)
You will see that this formula generates the numbers given in the sequence in its given position.
2)
In order to determine if a number, like 786432 appears in this geometric progression, let's plug it in for \(a_n\).
\(786432=3*4^{n-1}\) | First, divide by 3 on both sides. |
\(262144=4^{n-1}\) | It is not necessary to solve for n to determine whether or not this number appears in this sequence. All we need to determine is that the number is divisible by the common ratio, 4 in this case. This number certainly is since the last two digits, 44, is divisible by 4, so the entire number is divisible by 4. This is valid since we are determining that this number fits the set common ratio, which it does. |