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3 years ago

a crew workers paved 2/15 mile in 20 minutes. if they work at the same rate what portion of a mile will they pave in one hour?

Mathematics

1 answer:

ANEK [815]3 years ago

6 0

Since 20 min is 1/3 of a mile 20 min*3=1 hour. then u multiply 2/5 by 3 to get 6/5.

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Answer:

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3 years ago

Complete the following statement. If AB = BC and A,B,C are collinear, then b is the BLANK of AC

algol13

Answer:

hi, need help? Call me!! 650-983-3345

Step-by-step explanation:

4 0

3 years ago

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

Is 3 correct pls helppppppppppp

Nadusha1986 [10]

Answer:

yes it is you didn't even have to ask the question

Step-by-step explanation:

hope this helped :)

6 0

3 years ago

PLS HELP ME GRADUATE ILL MARK BRAINLY A-AND ILL GIVE U A COOKIE!!! PLS

snow_lady [41]

Answer:

Dilate circle A by a scale factor of 4.

Step-by-step explanation:

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3 years ago