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Nastasia [14]
2 years ago
6

5.) Write an inequality that represents each statement

Mathematics
1 answer:
alekssr [168]2 years ago
6 0

Answer:

give me your points

Step-by-step explanation:

c.

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if we have a geometric series with u1 = 2.1 and r= 1.06, what would the least value of n be such that Sn > 5543?
bulgar [2K]

Answer:

  88

Step-by-step explanation:

Write the expression for the sum in the relation you want.

  Sn = u1(r^n -1)/(r -1) = 2.1(1.06^n -1)/(1.06 -1)

  Sn = (2.1/0.06)(1.06^n -1) = 35(1.06^n -1)

The relation we want is ...

  Sn > 5543

  35(1.06^n -1) > 5543 . . . . substitute for Sn

  1.06^n -1 > 5543/35 . . . .  divide by 35

  1.06^n > 5578/35 . . . . . . add 1

  n·log(1.06) > log(5578/35) . . . take the log

  n > 87.03 . . . . . . . . . . . . . . divide by the coefficient of n

The least value of n such that Sn > 5543 is 88.

3 0
2 years ago
A function is defined as {(0, 1), (2, 3), (5, 8), (7, 2)}. Isaac is asked to create one more ordered pair for the function. Whic
svet-max [94.6K]

Answer:

2

Step-by-step explanation:

3 0
3 years ago
If sirius rises at 8:00 pm tonight, at what time will it rise tomorrow night, to the nearest minute?
madam [21]
We know that
The Sun makes a complete circle in the sky approximately every 24 hours, while the stars make a complete circle in the sky in 4 minutes less time, or 23 hours and 56 minutes

<span>Sirius is a star
so
</span><span>If Sirius rises at 8:00 pm tonight
then
tomorrow </span>will it rise  at ------> 8:00 pm -4 minutes----> 7:56 pm

the answer is
7:56 pm

6 0
2 years ago
8^1/3 divided by 5^-2
erastova [34]

Answer:

50

Step-by-step explanation:

8 0
2 years ago
If 30% of John's photographs were perfect, but 210 photographs were not perfect, then how many photographs were there in all?
loris [4]
If 30% were perfect, 70% were not perfect.  So we know that 70% of the total number of photographs is 210.

p(70/100)=210  multiply both sides by 100

70p=21000  divide both sides by 70

p=300

So there are 300 photographs in all.
8 0
2 years ago
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