Lut Q1 Aiu '0 4.76 Lct Q 4,75 1n(3 | |Q4). Thcu T =Ssin satisfies: #(1O0Q} (A) 0 <T<1 (B) (E) <T 4<T <6 (C) 2 < T < 3 (D) 3 < T < 4 ...

Use Table 7.9 . $$\begin{array}{c|c|c|c|c|c} \hline t & 0.0 & 0.1 & 0.2 & 0.3 & 0.4 \\ \hline g(t) & 1.87 & 2.64 & 3.34 & 3.98 & 4.55 \\ \hline t & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 \\ \hline g(t) & 5.07 & 5.54 & 5.96 & 6.35 & 6.69 \\ \hline \end{array}$$ Estimate $\int_{0}^{0.6} g(t) d t$ using the midpoint rule with $n=3.$

We need to evaluate the following piecewise function, so we have Q evaluated at zero, so zero is in our first Range. So we're going to use the top equation, so this is just equal to four because there's no X. To replace letter B E falls between six and seven. So that means we're going to have to use our second equation. So this becomes negative E plus nine. See, and falls between one and two. So one and two fits our first equation, so that means that this is equal to four and then letter D, letter M falls between one and two, which is in our first equation. So this is also four.

Were given this factor value function here. So it's tea and t squared plus one, and we first want to find the limit of rt as to you Purchase Negative three. So let's go ahead and do that so we could just go ahead, imply the limit to each of these positions here. So this is gonna be the limit as t approaches Negative three, no teeth and the limit as t a purchase. Negative three of T squared plus one. Now T and T squared plus one are continuous functions. And because their continuous, we could just go ahead and plug the limited directly. So that's gonna be negative. Three. And then we have three square or negative three squared is nine plus one that gives us so the limit of this function is going to be negative. Timpte. Now, what about our A negative three? Well, that means we're going to take negative three and plug it into here. So that's going to give us negative three and then negative three squared plus one. So negative three. 10 would be are of negative three. Now they're asking, Is Artie continuous at negative three And the answer is yes. And that's true because the limit of tea approaching negative three of our T is equal to our of negative three we get, actually say more. So this is going to be continuous for all real numbers outside of even just x zero to negative three. Because each of our components here are continuous functions. But the only ask about negative three now, they tell us to find what is our of t plus two minus rt. So just go ahead and write. Both of these also are t plus two is going to be t plus two, and we have t plus two squared plus one. And then our tea was t t squared. Plus what? So now remember, when we subtract vectors, we subtract them component wise, so we'd have t plus two minus t. That's just gonna leave us with to. And then Wolf, we expand this Ellis, you forgive us. T squared plus four t plus four to that would give us five there. And so now if we were to go ahead and actually subtract these use of the T Square's capsule and even so that cancels the one cancels and we're gonna be left with four t plus or so. This here is going to be the difference of those two

All right. If I was given Our beautiful function we've been working with here. So you are the nine and asked Which which methods underestimate this? Um Given that it's always increasing. First of all, I understand that if I do a left estimate, that one is definitely going to be lower than the right since I'm using the lower values. And since it is constantly increasing, finally take this chunk, it's going to be an under estimate of the total since my biggest values are not used. So one of them is definitely going to be a left handsome. It's going to be an underestimate. That's my first one. Okay. To make this a little bit more electable too. And some Okay, that's the first one. The right handsome. We definitely know it's going to be an overestimate and when it comes to mid point we know that can cavity is not changing. Um so midpoint is also going to be a underestimate. Yeah. Okay. And we we know that um yeah, my writing is just not good, right? We know that any time we set one, mid point is always the opposite of trapezoid. If midpoint is under then trapezoid is over. If trapezoid is over, the midpoint is under. So since mid point is under, that trapezoid is definitely over. So these are our two.

In this problem, we have been asked to find an exponential function that fits the data that were shown in our chart. Now, as a reminder, when we have an exponential function, it will take the form and I'm gonna use Y and T because those are variables in this problem. Why equals B times A to the t some constant times and number raised to the variable power. So let's see if we could figure out what A and B need to be in order to fit this data. To do that, we'll start with our first point. Our first point is when t is zero. Why is 600? So let's plug those in. Why is 600 and t is zero? Well, anything to the zero power is one. So a to the zero equals one, leaving us with B equaling 600 so I can plug that into my equation. I now have. Why equal 600 times age? The T. I just don't know a yet to do that. Let's go to our second point. Our second point is one 630 and let's plug that in. That gives me 630 equals 600 times a to the first. So all we need to do is divide by 600 to find a and A equals 1.5 Yeah. Now, the final form of my equation, why equals 600 times 1.5 to the T power. It's always good to try the other data points that we know to make sure this truly is a good fit for our data. So let's plug in some other values of T that we've been given. If I plug in t equals two into this equation, we do indeed get 661.5. Plugging in three gives us 694.58 Plugging in four gives us 729.3. It does match our data. This is a good fit for our our data that we've been given