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MatroZZZ [7]
3 years ago
13

Solve the following equation: lim x->0 sin3x/5x^3-4x.

Mathematics
1 answer:
goblinko [34]3 years ago
3 0
Usual limit of sin is sinX/X--->1, when X--->0

sin3x/5x^3-4x=0/0?, sin3x/3x--->1 when x --->0, so sin3x/5x^3-4x=                    [3x. sin3x / 3x] /(5x^3-4x)=(sin3x / 3x) . (3x/5x^3-4x)
   =(sin3x / 3x) . (3/5x^2- 4)
finally lim  sin3x/5x^3-4x=lim (sin3x / 3x) .(3/5x^2- 4)=1x(3/-4)= - 3/4
                 x----->0            x---->0
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1jaiz4 and 21 more users found this answer helpfulcandy sold vs money earned

money earned =  2 dollars * number of bars sold- cost of bars

money earned = 2 b -20

this is an equation of a line

(y= 2x-20)  

A.The relationship is linear    true

B.The relationship is nonlinear   false

C.The relationship is always results in a negative profit   false

it crosses the x axis at 10 candy bars

D.The relationship always result in a positive profit   false

it crosses the x axis at 10 candy bars

E.The relationship is proportional   false

F.The relationship is non-proportional  true  it is a line that does not cross the origin

3 0
2 years ago
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Answer:

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Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Simplify the complex fraction. Please show all work!
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Answer:

3(u^2 +u-3)

-------------------

u(u-3)^2

Step-by-step explanation:

u         1

---- + -------

u-3      u

-------------------------

    u-3

  ----------

    3

Get a common denominator for the numerator

u *u         1(u-3)

----      + -------

(u-3)u      u(u-3)

-------------------------

    u-3

  ----------

    3

u *u + (u-3)

-----------

(u-3)u      

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Now use copy dot flip

u *u + (u-3)        3

----------------- * -------------

(u-3)u                 u-3

3(u^2 +u-3)

-------------------

u(u-3)^2

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