Find the Volume of a cude with side lengths 3w^4 *

4. Your class is going to elect a President and Vice President. In how many ways can 2 people be elected if your class has 30 st

lina2011 [118]

Ok so all you have to do is

5 0

2 years ago

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MARKIN BRAINIEST!!! GIVIN 30PTS!!!

Sedbober [7]

Answer:

(-16x² +10x -3) +(4x² -29x -2)
(2x² -11x -9) -(14x² +8x -4)
2(x -1) -3(4x² +7x +1)

Step-by-step explanation:

I find it takes less work if I can eliminate obviously wrong answers. Toward that end, we can consider the constant terms only:

-3 +(-2) = -5 . . . . possible equivalent
-10 -5 = -15 . . . . NOT equivalent
3(-5) -2(5) = -25 . . . . NOT equivalent
-9 -(-4) = -5 . . . . possible equivalent
-7 -(-5) = -2 . . . . NOT equivalent
2(-1) -3(1) = -5 . . possible equivalent

Now, we can go back and check the other terms in the candidate expressions we have identified.

1. (-16x² +10x -3) +(4x² -29x -2) = (-16+4)x² +(10-29)x -5 = -12x² -19x -5 . . . OK

4. (2x² -11x -9) -(14x² +8x -4) = (2-14)x² +(-11-8)x -5 = -12x² -19x -5 . . . OK

6. 2(x -1) -3(4x² +7x +1) = -12x² +(2 -3·7)x -5 = -12x² -19x -5 . . . OK

All three of the "possible equivalent" expressions we identified on the first pass are fully equivalent to the target expression. These are your answer choices.

7 0

3 years ago

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Consider a melody to be 7 notes from a single piano octave, where 2 of the notes are white key notes and 5 are black key notes.

igor_vitrenko [27]

Answer:

35,829,630 melodies

Step-by-step explanation:

There are 12 half-steps in an octave and therefore $12^7$ arrangements of 7 notes if there were no stipulations.

Using complimentary counting, subtract the inadmissible arrangements from $12^7$ to get the number of admissible arrangements.

$\displaystyle \_\_ \:B_1\_\_ \:B_2\_\_ \:B_3\_\_ \:B_4\_\_ \:B_5\_\_$

$B_1$ can be any note, giving us 12 options. Whatever note we choose, $B_2, B_{...}$ must match it, yielding $12\cdot 1\cdot 1\cdot 1\cdot 1=12$ . For the remaining two white key notes, $W_1$ and $W_2$ , we have 11 options for each (they can be anything but the note we chose for the black keys).

There are three possible arrangements of white key groups and black key groups that are inadmissible:

$WWBBBBB\\WBBBBBW\\BBBBBWW$

White key notes can be different, so a distinct arrangement of them will be considered a distinct melody. With 11 notes to choose from per white key, the number of ways to inadmissibly arrange the white keys is $\displaystyle\frac{11\cdot 11}{2!}$ .

Therefore, the number of admissible arrangements is:

$\displaystyle 12^7-3\left(\frac{12\cdot 11\cdot 11}{2!}\right)=\boxed{35,829,630}$

6 0

2 years ago

Multiplying Polynomial

Alex Ar [27]

(2a+2)(4a+4)
8a^2+8a+8a+8
The answer is 8a^2+16a+8

5 0

3 years ago

The sum of the measure of two angles is 183.6 degrees. On angle has measure of 54 degrees. What is m, the measure in degrees of

Bas_tet [7]

Simple,

you have 2 angles...when added together they equal 183.6

The measure of one of those angles is 54°

What's the measure of the other angle?

183.6
-
54

=129.6

Thus, m=129.6, your answer.

8 0

3 years ago

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