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zavuch27 [327]
3 years ago
13

Help me pleasseeeee!!! its about algebraic properties of limits!

Mathematics
1 answer:
s2008m [1.1K]3 years ago
5 0

\text{Hello there! :)}

3. \sqrt{2} \\\\\text{4.  0}\\\\\text{5.  dne}

\text{To find the overall limit, they must approach the same y-value from each side:}

\\\\3.  \lim_{x \to \frac{\pi }{4} } 2sinx ={\sqrt{2}

\text{Therefore:}\\\\3.  \lim_{x \to \frac{\pi }{4} } f(x)= {\sqrt{2} }

\text{For this question, evaluate the limit from the RIGHT-HAND side:}\\\\4.   \lim_{x \to \pi +} f(x)\\\\ \text{Use the equation tanx to evaluate since it involves values greater than \pi:}\\\\

\lim_{x \to \pi +} tanx = 0\\\\\text{So:}\\\\ \lim_{x \to \pi +} f(x) = 0

5.  \\\\\text{Evaluate the overall limit. Make sure both sides approach the same y=value:}\\\\ \lim_{x \to \pi } 2cosx= -2\\\\ \lim_{x \to \pi } tanx = 0\\\\ \text{ Therefore:}\\ \lim_{x \to \pi } f(x) = dne \text{ (Does not exist})

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In the jone school library,10/20 of the computers have scanners.in simplest form, which fraction of the computers have scanners?
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3 years ago
The diagrams may not be drawn to scale. Find the value of y. AC is a straight line. А 56° yº B 123° су
Sliva [168]
<h3><u>Answer:</u></h3>
  • The answer is y = 67°.
<h3><u>Step-by-step explanation:</u></h3>
  • 56 + y + (180 - 123) = 180
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<h3><u>Conclusion:</u></h3>

Therefore, the answer is y = 67°.

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3 0
3 years ago
The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t 2 i - 4tj - 1k6 lb, and F
olga2289 [7]

Answer:

r=294.9m

Step-by-step explanation:

The forces on the particle are

W=mg\hat{j}\\F_{1}=52\hat{i}+6\hat{j}-2t\hat{k}\\F_{2}=5t^{2}\hat{i}-4t\hat{j}-1\hat{k}\\F_{3}=(5-2t)\hat{i}

Now , we sum all these forces to get the net force

F_{T}=W+F_{1}+F_{2}+F_{3}\\F_{T}=(52+5t^{2}+5-2t)\hat{i}+((6+6-4t)\hat{j}+(-2t-1)\hat{k}\\F_{T}=(57-2t+5t^{2})\hat{i}+(12-4t)\hat{j}+(-2t-1)\hat{k}\\

we can use the fact F=m*a and integrate the acceleration

a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]

and we evaluate in r(2) an we take the norm to obtain the distance

r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m

I hope this is useful for you

regards

8 0
4 years ago
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