All Answers

+130560

+8

So I guess you want to graph y = l 7 - x l ???

"y" will never be negative in this graph.

You have the first part correct. When x = 0, y will = l 7 l = 7

So we have the points (7, 0) and (0, 7)

Connect these two points. Now, pick a value for x > 7, say 9. Then l 9 - 7 l = 2. So the point (9, 2) will be on the graph. Draw a line connecting this point and (7, 0). There's your graph.

The easiest thing to see is that this is the graph of y = lxl shifted to the "right" by 7 units ....!!!! y = lx + 7l would shift the basic graph to the "left" by 7 units ....

Does that help??

CPhillOct 31, 2014

#2

0

i tried uploading a bigger pic twice, has it not worked?

GuestOct 31, 2014

#2

0

I guess that e constant

GuestOct 31, 2014

#1

+118725

0

I cannot see the picture and I have no idea what you are talking about. LOL

MelodyOct 31, 2014

#18

+1832

0

nice , but my question why do you put the angle ( x^2) not (x^2+10 )

xvxvxvOct 31, 2014

#5

+130560

0

Well...if I don't know much about them, someone on here probably does.....post your questions......

CPhillOct 31, 2014

#17

+1832

0

thank you melody for your comment

xvxvxvOct 31, 2014

#4

0

True! So it wouldn't matter too much if I embarrased myself some more! How are you with modulus functions?

GuestOct 31, 2014

#3

+130560

+5

LOL!!!...don't worry about it......remember, you're "Anonymous"......

CPhillOct 31, 2014

#2

+5

I just realised that 5 seconds after i posted the question and now I feel so stupid! Sorry, only just woke up! XD

GuestOct 31, 2014

#1

+130560

+5

We don't need much familiarity with logs or exponentials to do this one. We're just asking, "What exponent do I put on 2 to equal 2??" The answer is 1. This is because 2^1 = 2 ....Thus, x = 1......

CPhillOct 31, 2014

#16

+130560

+10

Here's my approach to (c)

lim x → ∞ [sinx^2 + 10] /[x^ + 10]

Splitting this up, we have

lim x → ∞ [sinx^2 ] / [x^ + 10] + lim x → ∞ [ 10] / [x^ + 10]

Notice that the second thing just evaluates to 0

For the first...we can do the Squeeze Theorem "thing" again

-1 ≤ sin x^2 ≤ 1

And dividing each thing by x^2 + 10, we have

-1 / x^2 + 10 ≤ sin x^2 / x^2 + 10 ≤ 1 / X^2 + 10

And because the two "outside" functions approach 0 as x approaches infinity, then the middle one is "squeezed" to this limit, too.

Thus

lim x → ∞ [sinx^2 ] / [x^ + 10] + lim x → ∞ [ 10] / [x^ + 10] =

0 + 0 = 0 !!!!

CPhillOct 31, 2014

#15

+118725

+10

Your answer for e is good 315.

My only criticism (and it may be just a person dislike) is this

If -3<x<4

then technically

3>-x>-4

this is true but I don't like the presentation. I think the little number should go first and that it should be rearranged and presented as

-4<-x<3

Maybe this is just a personal bias, another mathermatician might like to comment.

MelodyOct 31, 2014

#14

+1832

+5

yes thank you

xvxvxvOct 31, 2014

#7

+118725

+5

Thanks Chris,

I really like it when people ask for clarification.

It means that they are really trying to learn from our answers and that is what we want. :)

MelodyOct 31, 2014

#6

+130560

0

Thanks for that excellent answer, Melody.........I hope "Anonymous" understood it better after that...

CPhillOct 31, 2014

#3

+33666

+10

1. 3x^(2/3)+4x^(1/3)-4=0

Let u = x^(1/3). The equation above becomes

3u^2 + 4u - 4 = 0

This can be factored as

(3u - 2)(u + 2) = 0

so u = 2/3 and u = -2

Because u = x^(1/3) we must have x = u^3 so

x = (2/3)^3 = 8/27 and x = (-2)^3 = -8

2. (1+1/x)^2 - 6(1+1/x) + 8 =0

Let u = 1+1/x. The equation becomes

u^2 - 6u + 8 = 0

This can be factored as

(u - 4)(u - 2) = 0

so u = 4 and u = 2

Since u = 1 + 1/x we must have x = 1/(u-1) so

x = 1/(4 - 1) = 1/3 and x = 1/(2 - 1) = 1

.

AlanOct 31, 2014

#1

+118725

+10

$$\frac{(x^2+2xy+y^2)}{(x^2-y)}*\frac{(7x^2-3xy-4y^2)}{(4x^2-3xy-7y^2)}\\\\
=\frac{(x+y)(x+y)}{(x^2-y)}*\frac{(x+y)(7x-4y)}{(x+y)(4x-7y)}\\\\
=\frac{(x+y)(x+y)}{(x^2-y)}*\frac{(7x-4y)}{(4x-7y)}\\\\
=\frac{(x+y)^2(7x-4y)}{(x^2-y)(4x-7y)}\\\\$$

I don't know if it can be simplified any further.

MelodyOct 31, 2014

#4

+118725

0

Thank you :)

MelodyOct 31, 2014

#13

+118725

+15

Hi 315

c)

Consider

lim x tends to infinity of (some little positive number divided by x)

=a little positive number divided by a huge number

the limit will be 0.

like

9/1000000000000000000000000000000000000000000000000000000000

is a very tiny number and the more digits you add to the end of it the smaller and closer to zero it becomes.

Does that make sense?

MelodyOct 31, 2014

#12

+1832

+5

for e I do it like this

and melody , I dont understand c

xvxvxvOct 31, 2014

#3

+5

I note there is an example you did a while back followed by a, well written, annotated explanation from another member.

Binomial distribution with The cumulative distribution function (binomcdf)

http://web2.0calc.com/questions/airlines-often-sell-more-tickets-for-a-flight-than-there-are-seats-because-some-ticket-holders-do-not-show-up-for-the-flight-assume-that-an#r4

Binomial distribution with The cumulative distribution function

1 - binomcdf(48,0.25,18)

Number of trials 48

Probability of success on a single trial 0.25

Number of successes (x) 18

This discreet value is subtracted from one giving the Cumulative Probability of: P(X < 18)

=0.9816

The calculator give all values: The discreet, and the cumulative inclusive and exclusive of the discreet in both directions.

GuestOct 31, 2014

#328

+118725

+3

@@ End of Day Wrap Fri 31/10/14 Sydney, Australia Time 4:15pm ♬

Hi All,

There has been lots of great answers the last two days. These were presented by Geno3141, CPhill, Alan, Will85237, Unknown, Linarose, Samanthamunguia99, reinout-g, darkangelsaura16, stormegetton, ZeroThe Protector and Kitty<3. Thank you all.