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Keith_Richards [23]

4 years ago

9

On monday, mary had 23 bananas. On tuesday, her family ate half of them. On wednesday they ate half of the remaining bananas. If

they continue each day, on which day of the week will there only be one banana left

1 answer:

telo118 [61]4 years ago

5 0

Answer:

Friday.

Step-by-step explanation:

The number of bananas is reducing from 23 at a rate of 50% of the previous day's number.

Given that on Monday, Mary had 23 bananas and we are asked that on which day there will be only 1 banana will left.

Let after d days the remaining bananas will become one.

Therefore, $23[1-\frac{50}{100} ]^{d} = 1$

⇒ $(0.5)^{d} =\frac{1}{23}$

Taking log both sides we get

$d \log 0.5 = \log (\frac{1}{23} )$

⇒ d = 4.523 days.

Therefore, on Friday the remaining number of bananas will become 1. (Answer)

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Tcecarenko [31]

Answer:

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Step-by-step explanation:

Given

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3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

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Let's take the partials derivatives.

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$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

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We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

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$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

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3 years ago

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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iren2701 [21]

Answer:

B

Step-by-step explanation:

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3 years ago