NotThatSmart

First, let's combine all our like terms and move all the terms to oneside. We get

\(6x^{2}-21x-31=0\)

Now, we use the quadratic equation. We get that

\(x=\frac{-({-21})\pm \sqrt{{-21})^{2}-4\cdot {6}({-31})}}{2\cdot {6}}\)

Simplifying furher. We get 2 values for x We get

\(x=\frac{\sqrt{1185}+21}{12}\\ x=\frac{-\sqrt{1185}+21}{12}\)

Clearly, the first value of x is greater in this case, so we have

\(x=\frac{\sqrt{1185}+21}{12}\) as our final anwer. 

Thanks! :)

Sorry, there was a slight error in my math. 

We have angle EHA as \(128-68=60\) degrees. 

Therefore, our answer iss

\(60+52=112\) degrees. 

Still not sure if this is right...

Thanks! :)

I'm not sure if my logic is correct, but I'll give it a try. 

The point where the 3 altitudes of a triangle meet is the orthocenter. H is the orthocenter in this problem. 

Now, let's extend segment CH until it meets AB. Name this point F. 

CF is the altitude now. 

Let's note that \(\angle CFA = 90^\circ\) since CF is now the altitude. 

We can write the equation \(\angle ACF = 180 - 90 - 52\) through triangle CAF, so we have \(\angle ACF = 38\)

Through triangle EHC, we have \(\angle CHE = 180 - 38 - 90 = 52\). Since angle CHE is equal to angle FHB, angle FHB is also 52 degrees.

Now, we can write the equation \(\angle BAC + \angle CFA + \angle BEA + \angle EHF = 360\) since it forms a quadrilateral. 

Plugging in what we have, we get that

\(52+90+90+\angle EHF = 360 \\ \angle EHF = 128\)

Now, angle EHF and angle AHB share angle AHF. Since FHB is 52, we have that angle AHF is 120 - 52 = 68 degrees. 

EHA is now 128- 68 = 20 degrees. 

Thus, angle AHC is equal to 20 + 52 = 72 degrees. 

So our answer is 72. 

I'm not sure if this is right....

Thanks! :)

b) Now, we can plug in -3/2 for x into the equation \(\frac{6x^3+9x^2+4x+7}{2x+3}=3x^2+2+\frac1{2x+3}\), and determine if we achieve equal factors. 

If we do, then it is a solution. 

If not, then it isn't a solution. 

Thus, we have

\(\frac{6(-3/2)^3+9(-3/2)^2+4(-3/2)+7}{2(-3/2)+3}=3(-3/2)^2+2+\frac1{2(-3/2)+3}\)

Now, let's focus on the denominators of the equation. We have that

\(2(-3/2)+3 = -3+3=0\)

This means that the left hand side of the equation is undefined. 

Also, we also have

\(2(-3/2)+3=-3+3=0\) for the denominator in the right hand side, which is also undefined. 

Thus, -3/2 is not a solution since both sides become undefined when we plug it in for x. 

Thanks! :)

a) Setting these two equations to equal to each other, we have

\(\frac{6x^3+9x^2+4x+7}{2x+3}=3x^2+2+\frac1{2x+3}\)

Now, if we plug 10 into the equation and it returns equal values, then 10 is an answer. 

However, if the two equations don't equal each other, then we know that 10 is not an answer. 

Plugging in 10, we have 

\(\frac{6(10^3)+9(10^2)+4(10)+7}{2(10)+3}=3(10^2)+2+\frac1{2(10)+3}\)

Simplifying both sides, we get that

\(\frac{6000+900+40+7}{20+3}=300+2+\frac{1}{20+3}\\ 6947/23 = 302 + 1/23\\ 6947/23=6947/23\)

Thus, we can easily tell that the two equations are equal. 

Thus, 10 is defintely a solution!

Thanks!: )

The formula of a circle is in the form of

\(\left(x−a\right)^2+\left(y−b\right)^2=r^2\) where (a,b) is the center and r is the radius. 

First, let's move all the terms to one side. We get

\(4x^2-8x+4y^2+14y-12=0\)

Dividing everything by 4, we have

\(x^2-2x+y^2+7/2y-3=0\)

Completing the square for both x and y, we have

\(\left(x-1\right)^2+\left(y-\left(-\frac{7}{4}\right)\right)^2=\left(\frac{\sqrt{17}}{4}\right)^2\)

Thus, the radius is just 

\(\sqrt{17}/4\)

Thanks! :)

When we are trying to find the roots, we set q(y) to equal 0. 

Thus, we have

\(12y^3-28y^2-9y+10=0\)

Factoring the polynomial, we have

\(12y^3-28y^2-9y+10=0\)

Thus, we have

\(y=\frac{1}{2},\:y=-\frac{2}{3},\:y=\frac{5}{2}\)

Thanks! :)