PLZ HELP!!!

Mathematics

1 answer:

Fed [463]3 years ago

4 0

4 is missing I assume. But I could be wrong

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GIVING BRAINLIEST POINTS Graph the parabola: (x+2)^2= -12(y-1)

Nutka1998 [239]

Answer:

did you type the right question give me the correct one and i will help

Step-by-step explanation:

4 0

3 years ago

Question 9 and 10 please help it’s due today and I will mark you brainless

GREYUIT [131]

Answer: the first one you graph the points (-4,11) (-3,4) (-2,-1) (-1,-4) (0,-5) (1,-4) (2,-1) (3,4) (4,11), it is a function because each input has only one output. Then the second is not a function because it does not pass the vertical line test

Step-by-step explanation:

4 0

3 years ago

The manufacturing cost of a cell phone was $136. Sellers first marked up the cost by 25 percent. Sales were poor, so sellers dec

Nuetrik [128]

The correct answer is:
C-old price was $170; new price is $159.12

3 0

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 $u$ ,

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