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**MajikWaffle**- Replies: 0

Let A = 5 9

2 7

and B = 1 8

3 6

Find the standard matrix C of the linear transformation T(x) = B(A(x)).

And then

Determine which of the formulas hold for all invertible *n*×*n* matrices A and B

A. A5 is invertible

B. (ABA−1)8 = AB8A−1

C. (A+B)(A−B) = A2−B2

D. (A+A−1)3 = A3+A−3

E. A+In is invertible

F. ABA−1 = B

And finally

Which of the following linear transformations from R3 to R3

are invertible?

A. Projection onto the xz -plane

B. Projection onto the x -axis

C. Rotation about the z -axis

D. Identity transformation (i.e. T(v) = v for all v)

E. Dilation by a factor of 6

F. Reflection in the y -axis

>_> Too hard.

**MajikWaffle**- Replies: 0

I'm looking for freeware that could create box and whisker plots. I don't have excel so I can't really do it there...

I would prefer one where I just plug the list into it and it generates a horizontal box plot, so any suggestions? Thanks.

Multiply 56 by (3 + 1/8) then divide by 25.

It's cross multiplying.

a/b = c/d

ad = bc

**MajikWaffle**- Replies: 5

*If a 1-meter stick is cut into two independent random places, what are the odds that the pieces can for a triangle*

So that's the problem right. I ran about 5 million trials with a self made program and got a probability of .25

Is that correct, and how can you algebraically prove it?

AP Stats? Take your TI-83

(a) normalcdf(8.0, ∞, 8.1, .1)

(b) normalcdf(8.2,8.3,8.1,.1)

(c) invNorm(.025,8.1,.1)

or you could use the 68-95-99.7 rule.

Heh, the problem changed once he figured it was unsolvable including the linked image.

Now it's solvable...

Also, sudoku puzzles are impossible to solve based on the predefined numbers in boxes.

The old image was

_ 2 _ |1 5 _ |_ _ _

_ _ 5 |2 _ _ |_ _ 9

3 _ _ |_ _ 4 |_ 2 5

---------------------

_ 8 7 |_ _ 2 |5 _ _

_ 4 _ |_ 1 7 |_ _ _

_ _ _ |_ _ 3 |1 _ 7

----------------------

2 _ 4 |6 _ _ |_ _ _

6 _ _ |_ 9 _ |_ 3 _

_ 5 _ |_ _ _ |9 6 8

**MajikWaffle**- Replies: 9

Teacher gave that as a problem of the week, and from what I've done I figured it was unsolvable. Kind of messed up POW if it doesn't even work...

**MajikWaffle**- Replies: 3

The cost of four items was $7.11 whether you added or multiplied the individual items. What was the cost of each item.

Our problem of the week for precalc. Problem is whether there is a set number of solutions. So help me out if you can please

It's just a like quadratic.

x^4 - 3x^2 - 4 = 0

(x^2 - 4)(x^2 +1) = 0

x^2 - 4 = 0 x^2 + 1 = 0

x^2 = 4 x^2 = -1

x = +- 2 x = +- i

n <- end of the sequence

Σ (Equation)

k = 1 <- starting

**MajikWaffle**- Replies: 5

http://img228.imageshack.us/img228/1799/untitled8av.png

can someone explain this to me?

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