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elena-14-01-66 [18.8K]
4 years ago
6

What does 5 half pints equal to in cups

Mathematics
2 answers:
uranmaximum [27]4 years ago
7 0
5 customary cups is the answer
andrew11 [14]4 years ago
7 0
3/3 because there's 8 pints in one gallon and just draw the picture of a gallon may and draw a line through that
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what is an equation of the line that is parallel to the line with y=-.60x+2 and passes through the point (10,-10)
sp2606 [1]

y = -0.60x - 4

because the slope stays the same because it is parallel and the y intercept for the new equation ends up being -4

4 0
3 years ago
111=13, 113=35, 115=57, 117=?
Elden [556K]
In the pattern, the values are increasing by 22 each time so for the 117th term you take 57 + 22 = 79

So, 117= 79
3 0
3 years ago
The quantity of soap produced in 2009 is 3000bars. if this is a 20% increase of what was produced in 2008.what is the quantity o
mr Goodwill [35]
2,400
20% of 3,000= 600
3,000 - 600= 2,400
6 0
3 years ago
Please help, thank you
alexandr1967 [171]

Answer:

falseeeeeeeee for sure

Step-by-step explanation:

can i get brianlissssstttt

4 0
3 years ago
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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