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3 years ago

How much is 1*1 ?? and is it rite right

1 answer:

4 0

Answer:As an adjective, right means "righteous, correct, or appropriate"; as a noun, it means "that which is just or legal." Write means "to inscribe or compose." A wright is someone who builds or repairs something. Rite is a ceremonial ac

Step-by-step explanation:

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iVinArrow [24]

Answer:

I think its 1.5 because 3 divided by 3 is 1.5

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3 years ago

Statistics show that the sales force of Golden Wholesalers successfully closed 1,711 sales out of 1,950 sales calls. What was th

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1711 sales out of 1950 calls. The success rate is quite impressive: 1711/1950=87.7%

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4 years ago

The concentration of the substance outside the cell and C2 represents the concentration of the substance

BigorU [14]

Is this math? I'm sorry I don't get what formula I'm supposed to rewrite or -

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3 years ago

100 points please help !!!! right answer only

DochEvi [55]

$864 {in}^{2}$

is the answer of the question

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3 years ago

A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres

topjm [15]

Let $U(t)$ denote the amount of uranium-238 in the reactor at time $t$ . As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

$\dfrac{\mathrm dU}{\mathrm dt}=-kU$

where $k>0$ is constant. Separating variables and integrating both sides gives

$\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}$

Suppose we start some amount $u$ . This means that at time $t=0$ we have $U(0)=u$ , so that

$u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}$

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if $t$ is taken to be in years) for some starting amount $u$ ,

$0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}$

The half-life is the time $t_{1/2}$ it takes for the starting amount $u$ to decay to half, $0.5u$ :

$0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}$

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

7 0

3 years ago