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Galina-37 [17]
4 years ago
9

‼️correct answer will get brainliest‼️

Mathematics
1 answer:
sveta [45]4 years ago
3 0
48 qrts * 25 barrels per trip = 1200 qrts get picked up per trip

1200 qrts * 1 hr/7qrts = 171.43 hrs to fill all 25 barrels

171.43 hrs * 1 day/24hrs = 7.14 days
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Hi can someone help me please
tensa zangetsu [6.8K]

Answer: yes

Step-by-step explanation:

3 0
3 years ago
Add 8.563 and 4.8292. then times the answer by 9.183 divide it 2 times by 1.192. Add to the answer with this number after your f
mina [271]

Answer:

105.876373581

Step-by-step explanation:

8.563 + 4.8292 = 13.3922

13.3922 × 9.183 = 122.9805726

122.9805726 ÷ 1.192 = 103.171621309

103.171621309 ÷ 1.192 = 86.5533735811

86.5533735811 + 19.323 = 105.876373581

6 0
3 years ago
Read 2 more answers
The seventh grade enrollment at Main Middle School reduced from 615 students to 492 students. By What percentage was the number
Contact [7]

Answer:

492 *100/625 =80 %

Step-by-step explanation:

Where [*]= Multiply

4 0
3 years ago
Faelyn noticed that she does not have a common factor. Which accurately describes what Faelyn should do next?
Jlenok [28]
Without knowing what Faelyn has done so far or what she is working with, the second selection seems appropriate.

6x^4 -5x^2 -4
= (6x^4 +3x^2) -(8x^2 +4)
= 3x^2*(2x^2 +1) -4(2x^2 +1)
= (3x^2 -4)(2x^2 +1)
4 0
4 years ago
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A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th
77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

A=A_{o} e^{-kt}  (1)

Where:

A=7 g is the quantity left after time t

A_{o}=12 g is the initial quantity

t=70 min is the time elapsed

k is the constant of decay for the material

So, firstly we need to find the value of k from (1) in order to move to the next part of the problem:

\frac{A}{A_{o}}=e^{-kt}  (2)

Applying natural logarithm on both sides of the equation:

ln(\frac{A}{A_{o}})=ln(e^{-kt})  (3)

ln(\frac{A}{A_{o}})=-kt  (4)

k=-\frac{ln(\frac{A}{A_{o}})}{t}  (5)

k=-\frac{ln(\frac{7 g}{12 g})}{70 min}  (6)

k=0.00769995 min^{-1}  (7)  Now that we have the value of k we can solve the other part of this problem: Find the time t for A=2 g.

In this case we need to isolate t from (1):

t=-\frac{ln(\frac{A}{A_{o}})}{k}  (8)

t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}  (9)

Finally:

t=232.697 min \approx 233 min

5 0
3 years ago
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