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4 years ago

Please help!!!!!!!!!

Mathematics

1 answer:

NemiM [27]4 years ago

8 0

Answer:

2^4÷8×530[45690×5+3=120981238765

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If you rotate triangle ABC 90 degrees counter- clockwise, where would the vertices of A'B'C' be located?

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a at c place 45 degree b at a place 45 dgree cat b place 90 degree

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3 years ago

Determine if the following is a linear equation. If so, write the equation in standard form. y-3/4 = x/2

nikdorinn [45]

The answer is:
x/2-y=-3/4

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4 years ago

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The population of Bridgetown triples every decade. If the population

Afina-wow [57]

Between 2000 and 2030 three decades will have gone by, which means the population will triple three times.

$2000-25,000 \\ 2010-25,000*3=75,000 \\ 2020-75,000*3=225,000 \\ 2030-225,000*3=675,000$

Therefore, the population of Bridgetown will be 675,000 people in 2030.

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3 years ago

Determine how many, what type, and find the roots for f(x) = x4 + 21x2 − 100. Determine how many, what type, and find the roots

kykrilka [37]

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Please, see the attached files.

Step-by-step explanation:

Please, see the attached files.

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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.

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3 years ago