proyaop

c) In part b we showed how these two expressions were undefined when x = -3/2, but now we must show why the two expressions are the same for all x except that one case- one can do this through manipulation...

The numerator, 6x^3 + 9x^2 + 4x + 7 can be manipulated and grouped to (2x + 3)(3x^2) + (2x + 3)(2) + 1, and you can see why they are the same after distribution *all I did was take the Greatest Common Factor = 2x + 3 out of the polynomial after grouping terms*

Then: (6x^3 + 9x^2 + 4x + 7)/(2x + 3) = [(2x + 3)(3x^2) + (2x + 3)(2) + 1]/(2x + 3). The 2x + 3 coefficients cancel out and you are left with a remainder of 1:

[(2x + 3)(3x^2) + (2x + 3)(2) + 1]/(2x + 3) = 3x^2 + 2 + 1/(2x + 3), which is the exact same as the second expression!

b) Explain why these two expressions do not represent equal numbers when \(x = -{3\over{2}}\).

b) You can show this by plugging -3/2 in- but wait! Can the denominator be 0? No, then the expression would be undefined!

The denominator, 2x + 3, is 2(-3/2)+3 = -3 + 3 = 0 when x = -3/2... So, these two expressions, each undefined, would not and could not be equal.

\({6x^3+9x^2+4x+7\over{2x+3}}, 3x^2 + 2 + {1\over{2x+3}}\)

a) To show that the two expressions represent equal numbers when x = 10, substitute 10 for x into each expression:

(6 * 10^3 + 9 * 10^2 + 4 * 10 + 7)/(23) = (6000 + 900 + 40 + 7)/23 = 6947/23

3 * 10^2 + 2 + 1/23 = 302 + 1/23 = 302*23/23 + 1/23, add numerators only: 6946/23 + 1/23 = 6947/23, equivalent to the expression above.

All sides of PQRS are the same, so we know this quadrilateral must be a rhombus, which means that the diagonals are perpendicular and bisect each other.                        Both PR and QS are diagonals, with PR = QS = 8. Because these diagonals are the same length and PQRS is a rhombus, we know that PQRS is specifically a square.                                  If the side length of a square is x, then the diagonal is x * sqrt(2). So, our equation is\(x*\sqrt{2}=8, x={8\over{\sqrt{2}}}={8*\sqrt{2}\over{\sqrt{2}*\sqrt{2}}}={8\sqrt{2}\over{2}}=4\sqrt{2}\). Now we know the side length of PQRS is 4sqrt(2). The perimiter would be four times that.

So, the answer is 4 * 4sqrt(2) = \(16\sqrt{2}\).

:)

Both Sarah and Amos say 0.

Sarah says all multiples of 3, and Amos says all multiples of 6 (under 100), because 6 is a multiple of 3 (3 x 2), then all the numbers Amos says, Sarah will say too. So now the problem becomes how many multiples of 6 are under 100? 

The largest multiple of 6 under 100 is 96 (6 x 16), and the smallest would be 6 (6 x 1). 6 x 1 to 6 x 16 would yield 16 multiples.

So Amos says a multiple of 6 sixteen times, not including 0. Hence, 1 + 16 = 17 numbers that both Sarah and Amos say out loud.

Nothing scary in listing out group 4 and 5, as they are pretty small: shouldn't take long :)

Group 4: 26, 28, 30, 32, 34, 36

Group 5: 38, 40, 42, 44, 46, 48, 50

Sum of Group 5 = 38 + 40 + 42 + 44 + 46 + 48 + 50, pair up the units digits to make a 10 for ease of calculation:

38 + 42, + 44 + 46, + 40 + 50, + 48

38 + 42 = 80

44 + 46 = 90

40 + 50 = 90

48 = 48

80 + 90 + 90, + 48 = 260 + 48 = 308

Or just use a calculator to get 308 lol (although not encouraged)

Nice one Harry... I agree with CPhill! 

Let's list the 3-partitions of 8 out, since 8 is pretty small:

3-3-1-1

3-1-1-1-1-1

1-1-1-1-1-1-1-1                                             So in total there are THREE 3-partitions of 8.

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Now, if you are trying to find the # of 3-partitions of a number bigger than 8 (perhaps much larger), you can do casework (a strategy where you count the number of cases for each broad possibility) on the number of times the largest power of 3 appears, and then within that you can do another casework for the number of times the second largest power of 3 appears, so on so forth until you end up with 1s.

For example: # of 3-partitions of 28:

Case 1: One appearance of 3^3

27-1

Case 2.1: No 3^3, but two 3^2

9-9-3-3-3-1

9-9-3-3-1-1-1-1

9-9-3-1-1-1-1-1-1-1

9-9-1-1-1-1-1-1-1-1-1-1

Case 2.2: No 3^3, only one 3^2

9-3-3-3-3-3-3-1

Instead of listing out everything, note how a 3 can be expressed as 3 or 1-1-1, so you can either have six 3s, five 3s... 0 3s, total of 7 cases- a much better strategy and time-saver as to listing out all.

Case 3: No 3^3 or 3^2, only 3^1

3-3-3-3-3-3-3-3-3-1

You can have nine 3s, and a 1, eight 3s and four 1s... and counting the number of appearances of 3s, you can have nine appearances all the way to 0 appearances: 10 cases

Case 4: Only 1s.

Clearly, a 1-1-1-1-1-1... string of twenty-eight 1s.

In total, 1 + 4 + 7 + 10 + 1 = 23

Ooooo note how 28 - 5 = 23, and 8 - 5 = 3. Where you subtract 5 from the original number to get the number of 3-partitions... coincidence or not? >:) I leave you with this question parmen

Yeah, there are a lot of questions here, so I'll leave you with this:

For a sin function\(f(x)=a\sin{(bx-c)}+d\) with parent function f(x) = sin(x), meaning all forms of the actual function are just transformations on sin(x):

1. Amplitude = |a|, if a < 0, vertical reflection of f(x)

2. Period (which means how long before the graph repeats itself) = 2pi/|b|, if b < 0, horizontal reflection of f(x)

3. Phase shift of c/b, and note that horizontal translations are generally the opposite of the sign.

4. + d refers to the increase/decrease proportionate to d's value of the midline; a vertical shift of f(x).

You can double check your answers or redo them with this information! :D