<span>There are 4 long rows and 5 short rows in the theater. x represents the number of chairs in each long row and y represents the number of chairs in each short row.
So, total number of chairs in 4 long rows= 4x
Total number of chairs in 5 short rows = 5y
Total number of chairs in the theater on a normal day = 4x + 5y
When 2 chairs are added to each long row, the number of chairs will change to (x+2).
So, total number of chairs in 4 long rows will be = 4(x+2)
When 3 chairs are added to each short row, the number of chairs will change to (y+3)
So, total number of chairs in 5 short rows will be = 5(y+3)
Thus, total number of chairs in the theater in rush day = 4(x+2) + 5(y+3)
= 4x + 8 + 5y + 15
= 4x + 5y + 23
Thus we can say the number of chairs increase by 23 as compared to a normal day.
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20% of 32= 6.4
20% x 32
20 /100 x 32
Reduce the fraction
1 /5× 32
=32/5
=6.4
Answer:
8/12 left
Step-by-step explanation:
8/12 of a bag left
2 1/12 = 25/12
1 5/12 = 17/12
25-17 = 8
The answer is 8/12
Answer:
Check below, please.
Step-by-step explanation:
Hi, there!
Since we can describe eccentricity as
a) Eccentricity close to 0
An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)
b) Eccentricity =5
An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:
c) Eccentricity close to 1
In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.