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3 years ago

What is the volume of the sphere of 6 feet

Mathematics

1 answer:

zaharov [31]3 years ago

6 0

The answer is V= 4/3 pie r^3

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What is the value of y?

kirill115 [55]

We are going to need more contex to awnser this

5 0

3 years ago

PLESE HELPPP!!!!!!!!!!!!!!!!

zzz [600]

Answer:

B. $\frac{6}{2x^{2} - 5x}$

Step-by-step explanation:

The product of the ratioal expressions given above can be found as follows:

$= \frac{2}{x} * \frac{3}{2x - 5}$

Multiply the denominators together, and the numerators together, separately to get a single expression

$\frac{2(3)}{x(2x - 5)}$

$= \frac{6}{x(2x) - x(5)}$

$= \frac{6}{2x^{2} - 5x}$

The product of the expression $\ = \frac{2}{x}*\frac{3}{2x - 5}$ = $\frac{6}{2x^{2} - 5x}$

The answer is B.

8 0

3 years ago

Factor 10c^2-60cd +80d^2

ozzi

$10c^2-60cd +80d^2$

$(2c - 8d) (5c -10d)$

$this \\ is \\ because...$

$2c*5c=10c^2$

$2c * -10d=-20cd \\ -8d*5c=-40cd \\ -20cd-40cd=-60cd$

$-8d * -10d = 80d$

5 0

3 years ago

Factorize 2x^2-5/3xy-2y^2

vladimir1956 [14]

I hope it helps you

3 0

3 years ago

Read 2 more answers

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