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Let f(x) = 2x+1 x > 0_Use a graphing device to graph f (x): Label the axes appropriately. flo) 2+l 9+8-+=/ 1 +IO) {()-Z0t}-244+421.5Find the range of f (x): Kong...

Question

Let f(x) = 2x+1 x > 0_Use a graphing device to graph f (x): Label the axes appropriately. flo) 2+l 9+8-+=/ 1 +IO) {()-Z0t}-244+421.5Find the range of f (x): Konge (0,60)For which values of x does f (x) = 52 f(2) = 2(2)+ 4 2+(2) 22 = 4The half-life of C14 is 5730 years: If a sample of C14 has time 0, how long will it take until 10 grams are left? mass of 40 micrograms at 0n2 v lt) - Wb ext Wut) Lo2 48 573 Wb #0.OCOI20960 0.25 L= MkO

Let f(x) = 2x+1 x > 0_ Use a graphing device to graph f (x): Label the axes appropriately. flo) 2+l 9+8-+=/ 1 +IO) {()-Z0t}-244+421.5 Find the range of f (x): Konge (0,60) For which values of x does f (x) = 52 f(2) = 2(2)+ 4 2+(2) 22 = 4 The half-life of C14 is 5730 years: If a sample of C14 has time 0, how long will it take until 10 grams are left? mass of 40 micrograms at 0n2 v lt) - Wb ext Wut) Lo2 48 573 Wb #0.OCOI20960 0.25 L= MkO



Answers

Radioactive Decay Let $Q$ represent a mass, in grams, of carbon $14\left(^{14} \mathrm{C}\right),$ whose half-life 5700 years. The quantity present after $t$ years is given by $Q=10\left(\frac{1}{2}\right)^{t / 5700}$ a. Determine the initial quantity (when $t=0$ ). b. Determine the quantity present after 2000 years. c. Sketch the graph of the function over the interval $t=0$ to $t=10,000$

He himself, whereas to answer a question about radioactive decay. Using exponential functions and their graphs, we're told to let you represent a massive carbon 14 in grams. It's a radioactive element whose half life is 5750 years. We're told that the quantity of carbon 14 present after two years, is given by the model. Hugh equals 10 times, one half to be t over the half life 5007 to 15. The T is in years. Yeah, can't tell. In part they were asked to determine the initial quantity. This is when t equals zero, so q of zero. This is 10 times one half to be zero over 5715 which is 10 times one or 10, and this is in grants. Then in part B. We'll have to determine the quantity present after 2000 years. In other words, Q. Of 2000 and 10 times one half to the 2000 over 5715. This is approximately 7.846 for one hour. Dude, this is also in Grant Steve Every finally in part C rest to sketch the graph of this function over the interval T equals 02 t equals 10,000. Rifle of that right right there is going to sketch. We're not going to have an accurate drawing, but we should know a few points, for example. So then you get, like, sort of tweet Q of zero We already know has the 0.0 10. And then we should also have Roberts. Well, q of 10,000. That's cool. This is 10 times one half to the 10. Oh, over 5, 715. Columbus, Ohio is the place trade, huh? You motherfucker probably want to go be nice, Toyota. You know what I mean? You're, like show respect to the barely used. This is approximately 2973 They're gonna take away your franchise, and this is in grams. So we also have the point 10 0 to 973 So walk. I draw scale, right? I'll let X range from zero up to 10,000 and the y axis solve this range from zero up to 10. No, to mind is business when a future stuff him. Yeah. Yeah. So we have a point at 0 10 which is here, and at 10,002 0.973 This is about, uh, yes, here ish. And so the sketch of our graph looks failed. Something like this once. They just want to be killed. That is one that does the restaurant.

Okay. So here we have this function that gives how much radium is prison after two years. So our first step is to determine the initial quantity. So I went to equal zero. So how do we do that? The queue of zero. It's just going to be equal to 25 times one half To the 0/16. So zero anything to the zero with power is one. So our initial amount, It's just 25 g of radio. 25 g. That's our initial next Burt yeah. Okay, determine the quantity present after 1000 years. Okay, so one is queue going to be equal to what is Q. going to be equal to When T equals one 1000? What how do we figure this out? Well all we're going to do is if we have this equation Q equals 25. 1 half to the T. Over 16 100. Let me just double check that. That's right, Yep. What are we gonna do? We're gonna plug in this 1000 everywhere that we see A. T. So you should find that 20 Q. Is going to be called a. 25 times one half To the 1000 over 1600th power. And what do we get when you use your calculator to evaluate this? You should get 16 21 grams. Okay. 16.21 g. Okay. Our next step is to use a graphing utility to graph it over the interval 0 to 1000. So how do you do that? Well if you have your calculator all you have to do is plug dysfunction in to the Y. One equals button. So just plug it in 1/2 to the you use the little arrow key beneath the button. Clear to get the little um arrow key button and use an exponent. And then you hit graph and then you want to make sure your window your X window which is related to the T function here Goes from 0 to 5000. Mhm Okay. And you should see an exponential function coming down coming down super far. And it should look something like okay, first part C should look something like this. Not quite that fast and steep of course, but in that general shape, Okay, when will the quantity of radium B zero g? So based on this graph, we know that it's never going to be zero. Right, Never going to be zero. Now, why is that? Because we can see from this graph this will constantly approach zero. It will come very close as X increases but it will never actually reach zero. So you can plug a huge number in for X or T. And you'll find that it gets very close to zero but will never actually reach zero because because this is only having, because the factor is one half, this is exponential decay. If you think about it constantly multiplying by another set of 1/2 will make it smaller, Like 1/8 is much smaller than one half, and it will eventually send it very close to zero. But you can never get actual zero from multiplication of this because there's there's nothing you can get to be equal to zero here, you can't get at this equal to zero. Okay, so that's why it's never going to be zero because it's exponential decay that's never going to reach zero. There'll always be a small amount of it left. So that is how you solve this problem. To determine the initial quantity plug in T. Equals zero. That means zero years have passed. Then if we want to know how much is left after 1000 years, we just plug in this 1000 everywhere. We see a.t. Then we plug it into our calculators to graph it. And based on that graph and our knowledge of exponential functions, we know that the amount is never going to be zero.

Indiscretion. We have given function a T equal to a initial times. He to a public 80 for the A. We're looking for K. That's the unknown. So, fortunately, we have 5700 years as a condition, so we can replace the tea clerking 5700 years. So now we only have one in the show. So now we only have one unknown number, which is K, so we can cancel to a initial on both sides. So we have 82 power, 5700 years stunts. Okay, you go to one of two. He over half lying to both sides. This this dreaded re Why? Because we use a property here. Here we can use a property, which is lying. I need to the power X always equal to X like me up. So no, we can is really soft decay divided by 5700 on both sides, we have Kate equal to line one over to divide by 5700. We can put it in calculator to calculate a specific number. All we can rule riding into negative line two divided by 5700. We can also put it into the calculator. We get approximate number, which is negative. Stale point Zoo, zoo, zoo 1216 So we got an answer for K. So peculiar to ride a 80 function. So we have I have 80 equal toe initial A e to the power. Okay, which is negative. 0.1 to 1 60 So that's the answer for a Okay. Next question Were you got a function from previous answer? We have 80 function Now the question is asking when it's one over to initial times in Isha. So we can do here is make the whole function equal to 1/3 terms initial. Now we have the function to solve it. First, we can eliminate initial on both sides. Now we only have one of our ripple which is t We used the same property with crash and hey, so we have a line on both sides. Don't forgot the line. We can easily put the power down. So in order to get more space t equal to lying. One award three divided by the constant number on the left. So that's how we has night 1035 years. Approximately some tea. Go to 9035 years. Thanks for watching

Were asked to answer a question about where you act. Okay. Using exponential functions in the grass let you represent a massive radioactive 20 g was half light is 24,100 years before the quantity of plutonium present after two years is given by the model should go to Rome. Few equals 16 times one half to the T over that half life. 24,100 Exactly in part, they were asked to determine the initial quantity of plutonium. T zero while we have Q of zero is 16 times one half zero over 24,100. This is just 16 times one, which is 16. And the unit is in grants First day of spring. Yes, okay, then Part B were actually determine the quantity present after 75 thousands of years. In other words, we want to find Q of 75,000. Yes, the seasons. The same way. Let me miss in. This is equal through 16 times one half 75,000 sober 24,100. If you plug this into a calculator, this is approximately one point 85 mm. This is also in grams five devastated. Finally, in part C, we're asking a graph. Me, Tony, The graft dysfunction over the interval from T equals 02 t equals 150,000 over the domain of 150,000 years, which I was to do this I'm going to use Cosmos is free online. Graphing calculator. Here is the graph of the function from T equals zero up to 150,000. Notice the even. After 150,000 years, some of the radioactive plutonium still remains high school. Well, some one time.


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