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mel-nik [20]
2 years ago
7

Y=5+2x fill in the table using this function rule

Mathematics
1 answer:
Ratling [72]2 years ago
8 0

Step-by-step explanation:

hope it helps you..........

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2. Estimate the square root of the following
In-s [12.5K]

Answer:

1..9.9

2....23.9

3...106.9

4....21.8

7 0
3 years ago
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Will give BRAINEST
Svetlanka [38]
X-y=2 (*-2) --> -2x+2y=-4
2x+3y=14 
------------------
2x+3y=14 
 -2x+2y=-4
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5y=10
y=2


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3 years ago
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Félix grew 8 flowers with 2 seed packets. With 5 seed packets, how many total flowers can Felix have in his garden? Solve using
Crazy boy [7]

Answer:

20

Step-by-step explanation:

8/2 equals 4 and 5 x 4 = 20.

5 0
3 years ago
I will mark Brainliest to the most accurate answer... Please help)
Aleks [24]

Answer:

24

Step-by-step explanation:

12% of 200

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8 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
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