jfan17

Let a be the first term, and r be the rate of the arithmetic sequence. We have:

a + 4r = 9

a + 31r = -84

a + 22 r = ?

Realize that we can Subtract our first equation from our second equation. This gives us:

27r = -93

r = -93/27

since we already have that a+31r = -84, we can just subtract 9r from this equation to get:
 

a + 22r = -84 -(9*-93/27) = -84 -(-31) = -84 + 31 = -53

Let's first expand our LHS(left hand side).

We get:

\(2x^2-3x-35 = -43 + jx\)

Adding 43 on both sides,

\(2x^2-3x+8 = jx\)

Subtracting jx on both sides, 

\(2x^2-(3+j)x + 8 = 0\)

Next, realize that if we have exactly one real solution to this quadratic, that implies it must be a double root; in other words, two of the same real root(otherwise if it was one real and one imaginary, the quadratic wouldn't have real coefficients). In order for this to be true, the discriminant of our quadratic must equal 0.

Generally for quadratics, the discriminant is:

\(\sqrt{b^2-4ac}\)

In this case, our discriminant for this quadratic would be:

\(\sqrt{(3+j)^2-4*8*2} = 0\)

\(\)Squaring both sides, we get:
\((3+j)^2 - 4*8*2 =0\)

\(9+6j+j^2-4*8*2 = 0\)

\(9+6j+j^2-64 = 0\)

\(j^2+6j-55 = 0\)

Factoring this out, we get:
\((j+11)(j-5) = 0\)

Our two values for j that satisfy this equation are then -11 and 5. 

cis(x) = cos(x) + isin(x)

cis(11pi/6) = cos(11pi/6) + isin(11pi/6)

cos(11pi/6) = \(\sqrt3/2\)

sin(11pi/6) = 1/2

So the top then becomes:

\(6\sqrt5 *(\sqrt{3}+i)/2\)

This expands to:

\((6\sqrt{15} + 6i\sqrt{5})/2\)

Now we do the same with the denominator.

cis(pi/2) = cos(pi/2) + isin(pi/2)

cos(pi/2) = 0

sin(pi/2) = 1

cis(pi/2) = 0 + i

Our denominator then becomes:

\(3\sqrt6 * 0 + 3i\sqrt6\)

Our answer is then:

\(((6\sqrt{15} + 6i\sqrt{5})/2)/3i\sqrt{6} \) which ends up simplifying to:

\((6\sqrt{15} + 6i\sqrt5)/6i\sqrt6 \)

Rationalizing our denominator, we get:

\((6\sqrt{90} + 6i\sqrt{30})/36i = (3\sqrt{10} + i\sqrt{30})/6i\)

I posted part of the solution to this here:

https://web2.0calc.com/questions/help_17124

Remember, if the question asks you to put it in cis format, 

\(\)cis(x) = cos(x) + isin(x)

v = <-9, 2> is transformed using the transformation matrix A = [-4, 1, 1, 2] What is the resulting vector? Show all your work.

We have the vector <-9, 2>, so we just apply the matrix A to this vector. We get:

A<-9,2> = [-4, 1, 1, 2]<-9,2> = <-4*-9 + 1 * 2, 1 * -9 + 2 * 2 > = <36 + 2, -9+4> = <38,-5>

Sorry for the formatting, I am new around typsetting languages for mathematics like LaTeX so I'm still learning how to do things(like matrices :/) 

Algebra will be our friend in this problem. We already have our variables to work with, so let's set up some equations:

\(6^2 + b^2 = c^2\) by Pythagorean theorem

We also have that the perimeter is 18, meaning:

\(6+b+c = 18\)

\(b+c = 12\)

\(b = 12-c\)

Substitute back into our equation from the Pythagorean theorem, and we can solve for the value of c.

\(6^2 + (12-c)^2 = c^2\)

\(36 + 144-24c + c^2 = c^2\)

\(180 -24c = 0\)

\(24c = 180\)

\(c = 180/24 = 30/4 = 15/2\)

We have 9 "slots" or blanks we want to fill these 9 textbooks into. Here's a visual:

_ _ _ _ _ _ _ _ _  and each one can be thought of as holding one of these textbooks; be it biology or economics.

First, we want to consider ou restrictions; the problem says that there must be an economics textbook in the middle and a biology textbook at each end. We look to our first requirement, which is the economics textbook in the middle. It makes sense for us to fill in this spot first. We then have:

_ _ _ _ 4 _ _ _ _ ways for us to fill in the middle spot because there are 4 economics textbooks to choose from. Our next requirement or restriction is that there must be a biology textbook at each end. Let's then fill in these.

5 _ _ _ 4 _ _ _ 4, ways to do so, since in the first spot, there are 5 biology textbooks(out of 5) to choose from, and in the last slot, there are 4 left over(since you've picked one already).

After this, we have no more restrictions left. At this point, we have already placed 3 books, so that leaves us with 6 remaining. We can then fill out these slots however we want, which gives us:

5 6 5 4 3 2 1 4 number of ways to do these arrangements. Next, we just multiply all of these, which gets us:

5 * 6 * 5 * 4 * 3 * 2 * 1 * 4 = 14400 ways

Multiply both sides by (x+2) * (x+4) or "cross multiply" to get:

\(2x(x+4) = -6(x+2)\)

Expanding out, we get:

\(2x^2+8x = -6x - 12\)

Adding 6x and 12 on both sides of the equation, we then get the quadratic:

\(2x^2 +14x+12 = 0\)

We can then factor this into:

\((2x+2)(x+6) = 0\)

The values for which this is 0 are -1 and -6

I'm not too savvy with number theory, but maybe you could try using the euclidean algorithm with this problem?