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Vexierspiege
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Hi,

Okay, so in my math class they switched the curriculum to a “Discover-it-yourself” strategy, which means that the “text book” is just a bunch of questions and the questions try to “coax you through” discovering stuff. That’s really hard to do, and my teacher is always gone because she has Lupus :( so for the first time in my life I don’t understand what’s going on at all! Can anybody help me?

If anybody could answer any/all of the questions I can think of that I have right now, I would be REALLY appreciative. Thank you SO MUCH.

(1) How do you compute the limit on a graph like f(x) = (x^2 - 3x - 10) / (x + 2)? as X: approaches 0, approaches inifinity, approaches negative inifinity, and/or approaches 1?

(2)How do you simplify an expression to ax^n form, where a is a constant and x is a variable? The equations I have to simplify are: 7/x^2, 3 x square root of 4x and 5x / square root x.

Thank everybody so much for reading this! And helping me if you can :).

Thanks,
Nicole

This open post was written 1 year, 6 months ago | V/U/S: 324, 1, 2 | Edit Post | Leave a reply | Report Post

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Creal Default offline Verified User (1 year, 10 months) Long Term User Shouts: 2 #
Houston, TX, US | 1 year, 6 months ago (17 hours, 48 minutes after post)

The first one sounds like calculus. I’m so, so at it, so I’m sorry if my explaination isn’t clear.

“(1) How do you compute the limit on a graph like f(x) = (x^2 - 3x - 10) / (x + 2)? as X: approaches 0, approaches inifinity, approaches negative inifinity, and/or approaches 1?”

Ok, first we can factor the numerator into (x-5)*(x+2)

This will become ((x-5)*(x+2)) / (x+2)
And the (x+2) will cancel, leaving you with: x-5

As x approaches 0, the limit is -5.
As x approaches inifinity, the limit is inifinity.
As x approaches negative infinity, the limit is negative infinity.
As x approaches 1, the limit is -4.

“(2)How do you simplify an expression to ax^n form, where a is a constant and x is a variable? The equations I have to simplify are: 7/x^2, 3 x square root of 4x and 5x / square root x. “

Ok, for the first one x is in the denominator. A negative exponent will be able to change it to the form you want. In this case: 7*x^(-2)

Square roots are the same as an exponent raised to the half. I’m not sure how the original problem was written exactly, but I am assuming you meant this formula:
((3x)*(4x + 5x)^(1/2)) / (x^(1/2))

If this is correct, you can square 3x (= 9(x^2)) and distribute that into the square root: (36x^2 + 45x^2)^.5 / x^.5

You can then change the equation to where the whole fraction is under the square root: ((36x^2 + 45x^2) / x)^.5

Then divide by x to get (36x + 45x)^.5
And then add the two x’s: (81x)^.5
Lastly, you can facor out a 9 to get 9*(x^.5)

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