Xie Xie Niii

There are a lot of websites for men giving tips on where/how to hook up with women in DR.

Any tips for American women looking to hook up with guys in DR? Please, no judgmental responses.

I'm dating a Dominican guy, who just texted me something like "I'm so exhausted! :("

Is an appropriate response to this "awww.. pobrecito?" (Poor baby?)

I'm not Spanish-speaking, so I'm not sure when to use this phrase.

I'm dating a Dominican guy, who just texted me something like "I'm so exhausted! :("

Is an appropriate response to this "awww.. pobrecito?" (Poor baby?)

I'm not Spanish-speaking, so I'm not sure when to use this phrase.

What are the intervals that are increasing or decreasing, and what are the critical numbers? I think I did the derivative incorrectly.

I got that the first derivative is -4x/(x-2)^3, but I'm confused about how to find the intervals that are increasing or decreasing, and the max/min.

The solution says the derivative is -(x^2+9)/(x^2-9)^2 but I think this is incorrect.

Shouldn't it be -2x^2/(x^2-9)?

Completely forgot how to do this. We did not learn L'Hospital's rule yet. We're supposed to factor out the highest power of x in both the numerator and denominator, but I don't know how to factor x^2?

I also need to set them to 0 to get the critical values.

I got -2x/(x^2-1)^2 for the first derivative. Is this correct?

I believe there is a vertical asymptote at x=1, and a horizontal asymptote at y=1. Is that right?

I also need help finding the intervals of concavity and the inflection points.

Thank you!

Is this equivalent to infinity over infinity, which equals 1?

What about for negative infinity? Also equal to 1?

When I entered the equation 1/(x+1)^2 into a derivative calculator, it told me the derivative should be 1/(x^2 + 2x +1), but I don't think I got that result.

Did I do something wrong with my calculation? I've linked an image to my calculations, below.

http://imgur.com/EwITI

Thank you!

Am I supposed to factor out the highest power, x^4?

How do I set up cuberoot(x) = 1 - x equal to zero? I don't want to SOLVE for zero, I just want to set the equation to zero. (The problem is to use the Intermediate Value Theorem to prove there is a root.)

Are cuberoot(x) - 1 + x = 0

and

1 - x - cuberoot(x)

equal to each other?

Because when I plug in 0 for the first equation I get -1, but in the second equation I get 1. I thought these two equations were equal? In the second equation, I simply moved over cuberoot(x) to the right and made it -cuberoot(x). Did I do some wrong arithmetic here?

The answer key says I was supposed to move 1 - x to the left side of the equation and make it -1 + x.

16 minutes ago - 4 days

Is this equal to infinity? I know that x→5⁻ means "approaching x from the left" but does that change the way I have to solve this problem? What would the answer be if there was no minus sign next to 5?

Are cuberoot(x) - 1 + x = 0

and

1 - x - cuberoot(x)

equal to each other?

Because when I plug in 0 for the first equation I get -1, but in the second equation I get 1. I thought these two equations were equal? In the second equation, I simply moved over cuberoot(x) to the right and made it -cuberoot(x). Did I do some wrong arithmetic here?

The answer key says I was supposed to move 1 - x to the left side of the equation and make it -1 + x.

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

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

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I'm confused about the "specified interval." In class we were given two different x-values and had to plug them into the equation to see if the function crossed over the x=0 line. I don't understand what (1,2) is asking me to do. Are 1 and 2 different x-values? Or is that an (x, y) coordinate?

Thanks in advance.

I can't tell if the first e-mail sounds too curt, or the second and third e-mails sound too chummy. Or are all three acceptable? I am on formal terms with my advisor, however it's been 7 months since I last met him so I thought some kind of greeting might be polite. But if it sounds too friendly/unprofessional I don't mind leaving the greeting out.

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(1) Dear Dean X,

In the Fall 2011 semester I took a medical leave of absence from the [XYZ] program. Since that time I have been undergoing outpatient therapy, and there has been some improvement regarding the conditions that had been disruptive to me and my studies in 2011. I am still in ongoing therapy now, however my treatment team and I now feel that conditions have gotten better to the point where I can once again resume my pre-medical studies. May I start the petition to reenroll for the Fall 2012 semester at this time?

Best regards,

ABC

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(2) Dear Dean X,

Hope you are having a pleasant summer. In the Fall 2011 semester I took a medical leave of absence from the [XYZ] program. Since that time I have been undergoing outpatient therapy, and there has been some improvement regarding the conditions that had been disruptive to me and my studies in 2011. I am still in ongoing therapy now, however my treatment team and I now feel that conditions have gotten better to the point where I can once again resume my pre-medical studies. May I start the petition to reenroll for the Fall 2012 semester at this time?

Best regards,

ABC

====

(2) Dear Dean X,

Hope you are well. In the Fall 2011 semester I took a medical leave of absence from the [XYZ] program. Since that time I have been undergoing outpatient therapy, and there has been some improvement regarding the conditions that had been disruptive to me and my studies in 2011. I am still in ongoing therapy now, however my treatment team and I now feel that conditions have gotten better to the point where I can once again resume my pre-medical studies. May I start the petition to reenroll for the Fall 2012 semester at this time?

Best regards,

ABC

G(x) = ³√(x) * (1 + x³)

Because ³√(x) is a root function? What about (1 + x³)?

I multiplied the numerator by 4x/4x and the denominator by 6x/6x. My answer came out to be 4/6. Is this correct?

I know you have to multiply this by [√(1 + h) - +1] / [√(1 + h) - +1], but then after that I'm lost.