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3 years ago

Write an addition equation that can help you find 9-6. Explain your answer.

Mathematics

2 answers:

Oksana_A [137]3 years ago

8 0

6+3= 9 that is what I would do

Rainbow [258]3 years ago

4 0

6+3=9 this what i had to do to check me answers

please can i have a brainliest

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2 years ago

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).

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3 years ago