AdamTaurus

1) To find a perpendicular line, the slope is the negative reciprocal.

Get the equation into y=mx+b form.

5x+8y=-9

Subtract 5x from both sides.

\(8y=-5x-9\)

Divide both sides by 8.

\(y=-\frac{5}{8}x-\frac{9}{8}\)

Take the negative reciprocal of m and use it with (10,10) in point-slope form, y-y₁=m(x-x₁).

\(y-10=\frac{8}{5}(x-10)\)

Multiply the 8/5 through and add 10 to both sides.

\(y=\frac{8}{5}x-16+10\)

\(y=\frac{8}{5}x-6\)

Then, m+b would be 8/5-6. This is 1.6-6, or -5.6.

2) The slope of a line is \(\frac{y_2-y_1}{x_2-x_1}\).

Plug in the points (1,2) and (7,4)

\(\frac{4-2}{7-1}=\frac{2}{6}=\frac{1}{3}\)

So the slope is 1/3. Now use point-slope form.

\(y-2=\frac{1}{3}(x-1)\)

Distribute 1/3 and add 2 to both sides.

\(y={1\over3}x-\frac{1}{3}+2\)

Combine.

\(y=\frac{1}{3}x-\frac{1}{3}+\frac{6}{3}\)

\(y=\frac{1}{3}x+\frac{5}{3}\)

3) Put 3y+2x=7 into y=mx+b form.

\(3y=-2x+7\)

Divide by 3.

\(y=-\frac{2}{3}+\frac{7}{3}\)

The slope of a line perpendicular to another is the negative reciprocal of the first slope.

\(m=\frac{3}{2}\)

So 3/2 is the answer to #3.

1) f(x)= x⁴+5x³-x²-5x

Statements 1, 3, and 4 are correct. 2 is not because f(5) does not equal 0.

The Rational Root Theorem is:\

Let f(x) be a polynomial with integral coefficients. The only possible rational zeros of f(x) are: 

\(\frac{p}{q}\)

where p is a divisor of the constant term and q is a divisor of the leading coefficient.

So, fators of the constant on top and factors of the LC on bottom.

\(\frac{\pm1, \pm2, \pm4}{\pm1,\pm2}\)

Now, pair them together.

Divide all of the top factors by 1 first, then 2.

\(\pm1,\pm2,\pm4,\pm \frac{1}{2}, \pm1,\pm2\)

Repeating factors can be dropped so the possible factors are \(\pm1,\pm2,\pm4,\pm \frac{1}{2}\) or \((x+1),(x-1),(x+2),(x-2),(x+4),(x-4),(2x+1),(2x-1)\).

So your possible factors are correct.

To see which ones are factors, I woud graph it.

This graph shows that \((-2,0), (-\frac{1}{2},0), (2,0)\) are the zeros. (Here is the graph: https://www.desmos.com/calculator/fx9umcmzu0)

As factors those are (x+2), (2x+1), and (x-2).

Whatever the highest degree is in a polynomial is the number of complex factors.

Therefore, in this equation there are 4 real complex numbers.

1) For the first, the number of complex zeros you have is always the same as your highest degree.

So, for that one it would be 4 complex zeros.

2) f(x)=x⁴+3x²−4

Here is the graph: https://www.desmos.com/calculator/zm7f4iwvrb

From the graph, the two zeros are (-1,0) and (1,0).

Now, I'm going to use long division for these.

Pull off all the coefficients (Don't forget the missing terms. This problem is really x⁴+0x³+3x²+0x-4)

I'm gong to use (-1,0) first, since it makes x+1 as a factor.

(Sorry if the picture is a bit fuzzy, I tried my hardest to get a clear one.)

(x+1)(x-1)(x²+4)

Now plug the x^2+4 into the quadratic formula.

In x²+4, a=1, b=0, and c=4.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}={-0 \pm \sqrt{0^2-4(1)(4)} \over 2(1)}\)

Now simplify. -0 can just go away and so can 0^2.

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

4*1*-4=-16

\(x={\pm \sqrt{-16} \over 2}\)

The square root of 16 is 4 and the square root of -1 is i.

\(x=\frac{\pm4i}{2}\)

This can be simplified to \(x=\pm2i\).

So, the factors are \((x+1)(x-1)(x+2i)(x-2i)\).

Or in polynomial form it would be \((x-2)(2x^2-x-2)+3\).

Thanks!

The point of the joke is that the kid has 6 fingers on one hand, therefore he is hiding his hand in his pants pocket.

To find the slope of a line, the equation is \(\frac{y_2-y_1}{x_2-x_1}\).

Plug in the point values.

\(\frac{-1+13}{6+6}\)

Simplify.

\(\frac{-1+13}{6+6}=\frac{12}{12}=1\)

So the slope is 1. Now use point-slope form, which is \(y-y_1=m(x-x_1)\).

Now, it doesn't matter which point you plug in. I'll use (-6,-13).

\(y+13=1(x+6)\)

Distribute the 1.

\(y+13=x+6\)

Subtract the 13 from both sides.

\(y=x-7\)

Are you calling me poor? (Jk, good joke though.)

Since we're going for long jokes:

An engineer dies and reports to the pearly gates. St. Peter checks his dossier and says, "Ah, you're an engineer — you're assigned to h**l."

So the engineer reports to the gates of h**l and is let in. Pretty soon, the engineer gets dissatisfied with the level of accommodations and starts designing and building improvements.

After a while, they’ve got air-conditioning and flush toilets, escalators, elevators and so on ... and the engineer is a pretty popular guy.

One day, God calls Satan on the telephone.

"So, how's it going down there in h**l?" God says.

"Hey, things are going great. We've got air-conditioning and flush toilets and escalators. There's no telling what our engineer is going to come up with next!" Satan says.

"What? You've got an engineer? That's a mistake - he should have never gotten down there. Send him back immediately!" God says.

"No way! I like having an engineer on the staff - I'm keeping him!" Satan says.

"Send him back up here or I'll sue!" God says.

Satan laughs crazily and answers:

"Yeah, right. And just where are you going to get a lawyer?"