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Alexus [3.1K]
3 years ago
6

How to solve this problem I don’t get it at all

Mathematics
1 answer:
Zarrin [17]3 years ago
8 0
Use a calculator to help u and u will get your answer
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Evaluate the function at the given numbers( correct to six decimal places). Use the results to guess the value of the limit or e
netineya [11]

Answer:

Value of the limit is 0.5.

Step-by-step explanation:

Given,

F(x)=\frac{e^x-1-x}{x^2}

When,

x=1,F(1)=frac{e^1-1-1}{1}=e-2=0.718281

x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885

x=0.1, F(0.1)=\frac{e^0.1-1-0.1}{(0.1)^2}=0.517091

x=0.05, F(0.05)=\frac{e^0.05-1-0.05}{(0.05)^2}=0.508438

x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)

Correct upto six decimal places.

Now,

\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}   (\frac{0}{0}) form, applying L-Hospital rule that is differentiating numerator and denominator we get,

\lim_{x\to 0}F(x)

=\lim_{x\to 0}\frac{e^x-1}{2x}    (\frac{0}{0}) form.

=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)

Limit exist and is 0.5. That is according to (1) we can see as the value of x lesser than 1 and tending to near 0, value of the function decreases respectively. And from (2) it shows ultimately it decreases and reach at 0.5, consider as limit point of F(x).  

8 0
3 years ago
THIS TABLE LINEAR OR NON-LINEAR?
defon

Answer:

yes

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Solve the system by substitution. Check your solution.
zvonat [6]
B. (9,126)

<span>y + 18 = 16x
=>y=16x-18 
0.5x + 0.25y = 36 (multiply both sides by 4)
=>2x+y = 144
Substitute y=16x-8

=>2x+16x-8=144
=>18x=152
=>x=152/18=9 
y=16x-18
=>y=16(9)-18
=>y=144-18=126 
Answer: x=9 and y=126</span>
4 0
3 years ago
Thirty
11111nata11111 [884]

Answer:

djriztos5islrula4p

Step-by-step explanation:

7epz7epsurs8tfoy

6 0
3 years ago
Read 2 more answers
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
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