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Genrish500 [490]
2 years ago
11

What is this? k/8-5≥-6

Mathematics
1 answer:
ddd [48]2 years ago
4 0

Answer:

k≥-8

Step-by-step explanation:

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A certain substance doubles its volume every minute at 9 AM a small amount is placed in a container ,at 10 AM the container was
Y_Kistochka [10]

Answer:

the container is 1/4 full at 9:58 AM

Step-by-step explanation:

since the volume doubles every minute , the formula for calculating the volume V at any time t is

V(t)=V₀*2^-t , where t is in minutes back from 10 AM and V₀= container volume

thus for t=1 min (9:59 AM) the volume is V₁=V₀/2 (half of the initial one) , for t=2 (9:58 AM) is V₂=V₁/2=V₀/4  ...

therefore when the container is 1/4 full the volume is V=V₀/4 , thus replacing in the equation we obtain

V=V₀*2^-t

V₀/4 = V₀*2^-t

1/4 = 2^-t

appling logarithms

ln (1/4) = -t* ln 2

t = - ln (1/4)/ln 2 = ln 4 /ln 2 = 2*ln 2 / ln 2 = 2

thus t=2 min before 10 AM → 9:58 AM

therefore the container is 1/4 full at 9:58 AM

8 0
2 years ago
What is 10+2<br> just trying to give people more points!!
Sergio039 [100]

Answer:

12

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
2 years ago
What is the value of x?<br> 33<br> 65<br> C=?
loris [4]

Answer:

56

Step-by-step explanation:

8 0
2 years ago
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I need help wit 5/2 divided by 5/7 =
Agata [3.3K]

Answer:

7/2

Step-by-step explanation:

8 0
3 years ago
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