Question

6x^3 + 6x^3 ( 6x^3) + 6x^3 + x^2(x^4)

Answer 1

[-3\[[-3^{0}+2^{-1}-4^{-1}]^{-2}\]

Answer 2

ignore the [-3*

Answer 3

I need help on these 2 problems

Answer 4

@jazzylj --> We're ready to begin - waiting for you.

Answer 5

@oholmes I can't read the second one - equation editor mumbo jumbo on my computer.

Answer 6

[-3^0+2^-1-4^-1]^-2

Answer 7

6x^3 + 6x^3 ( 6x^3) + 6x^3 + x^2(x^4) = 37 x^6 + 12 x^3
------------
@jazzylj @oholmes What did you get?

Answer 8

overall I got 84x^3+x^6

Answer 9

6x^3 + 6x^3 ( 6x^3) + 6x^3 + x^2(x^4) = 37 x^6 + 12 x^3
I got the same thing

Answer 10

okay

Answer 11

do we add the 37 and 12?

Answer 12

Do you know how we got that?

Answer 13

yes I worked it out again

Answer 14

was there another problem?

Answer 15

6x^3 + 6x^3 ( 6x^3) + 6x^3 + x^2(x^4) =
6 x^3 + 36 x^6 + 6x^3 + x^6 =
37 x^6 + 12x^3 --> @oholmes @jazzylj Agree?

Answer 16

yes, agreed

Answer 17

yes @jazzylj its above 1st comment

Answer 18

yes

Answer 19

[-3^0+2^-1-4^-1]^-2
------------------------
What about this? --> 16/9 Agree?

Answer 20

anything to the zero power is 1.

Answer 21

yes 16/9. :)

Answer 22

@oholmes --> What did you get as an answer?

Answer 23

-3^0=1
2^-1=1/2
-4^-1=-1/4

Answer 24

16/91

Answer 25

almost...just without the 1 in the 91 :)

Answer 26

*9!

Answer 27

What next? Another question, @oholmes? If so, post in a new thread.

Answer 28

@Directrix...can you help me with limits?

Answer 29

@jazzylj --> I'm a little fuzzy on limits but I may be able to help.

Answer 30

Go ahead and post the question. @jazzylj

Answer 31

lim x-> infinity (x +sinx)/(2x+7-5sinx)

Answer 32

@jazzylj Did you get 1/2 (using L'Hospital's Rule, I think)

Answer 33

I think I got 0...but I'm really not sure

Answer 34

lim x-> infinity[ (x +sinx)]/[(2x+7-5sinx)] =

This is an 00/00 situation so apply L'Hospital's Rule.

lim x-> infinity [ 1 + cos x ) ] / [2 + 0 - 5 (cosx)]

@jazzylj --> Do you agree so far?

Answer 35

Yes

Answer 36

Hey, here's a good study/review source for the rule:
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/statement.html

Okay, I'll continue my work.

Answer 37

thank you @Directrix!

Answer 38

lim x-> infinity [ 1 + cos x ) ] / [2 + 0 - 5 (cosx)] =

lim x-> infinity [ 1 + cos x ) ] / lim x -> infinity [ [2 + 0 - 5 (cosx)] =

into parts:
Numerator: lim x-> infinity [ 1 + cos x ) ] = 1 + 0 = 1.
Denominator: lim x -> infinity [ [2 + 0 - 5 (cosx)] = 2 -5(0) = 2.

Back Together:

lim x-> infinity [ 1 + cos x ) ] / lim x -> infinity [ [2 + 0 - 5 (cosx)] = 1/2.

Answer 39

See if you think that is correct. As x goes gargantuan, cosx goes to 0.