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by maggiemagnet
Fri Jan 16, 2015 3:18 pm
Forum: Trigonometry
Topic: angles of a oblique triange
Replies: 4
Views: 348

Re: angles of a oblique triange

I think the usual way of saying it would like something like this:

"ABC is a triangle. Opposite the vertices A, B, and C are the sides a, b, and c, respectively. The lengths of the sides are: a = 88, b = 72, and c = 110. Find the measures of the angles at A and at C."
:clap:
by maggiemagnet
Fri Jan 16, 2015 3:14 pm
Forum: Beginning Algebra
Topic: Domain and Range of a relation
Replies: 5
Views: 526

Re: Domain and Range of a relation

Then which is the independent variable and which is the dependent variable? The domain and range depend on this information! Thanks!
:clap:
by maggiemagnet
Thu Jan 15, 2015 12:27 pm
Forum: Intermediate Algebra
Topic: Pondering Base and Exponent, Both NEGATIVE
Replies: 3
Views: 338

Re: Pondering Base and Exponent, Both NEGATIVE

I wrote up a problem for myself to test the behavior of a negative exponent with a negative base.... y\, =\, \dfrac{(-5)^{-x}}{34} Before I proceed, I will do a mini-proof of the numerator. (-5)^{-x}\, =\, \dfrac{x}{(-5)^1}\, =\, \dfrac{x}{(-5)} Is this math correct?...
by maggiemagnet
Sat Nov 22, 2014 2:03 pm
Forum: Statistics
Topic: I need help to solve this question
Replies: 1
Views: 510

Re: I need help to solve this question

The following table represents the absent days of a sample of students in the statistics course Absent days 0 1 2 3 4 Number of students 10 5 8 Y 3 Determine the value of Y if the median of the absent days is 1.5 day. The " median " is the middle value in the list. So write out the list: ...
by maggiemagnet
Mon Nov 10, 2014 6:29 pm
Forum: Statistics
Topic: Expected Value of Probability Density Function
Replies: 1
Views: 413

Re: Expected Value of Probability Density Function

I am trying to find the Expected Value of the following Probability Density Functions (where E is Euler's number) for the x values shown below. For some reason I get very different values from my calculations of the expected value when applying the discrete and continuous method of calculating expe...
by maggiemagnet
Sun Nov 02, 2014 12:18 am
Forum: Intermediate Algebra
Topic: beginner f(x) domain
Replies: 1
Views: 444

Re: beginner f(x) domain

I have f(x)= x^2 +5x +6 find the domain , all possible values of x The "domain" is all values of x that will work in the function. so f(x) = x^2 + 5x +6 factors out to (x+3) (x+2) Imy anser is x is all rational numbers except -3 or -2 Why is your function not allowed to equal zero? Dividi...
by maggiemagnet
Thu Oct 23, 2014 11:15 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: degree of quotient of polynomials
Replies: 1
Views: 474

Re: degree of quotient of polynomials

1) Is it correct to assert that the degree of the quotient is always equal to the degree of the numerator minus the degree of the denominator? I've never before heard of the "degree" of a rational function but, yes, I think it would be computed the way you've described. On the other hand,...
by maggiemagnet
Mon Sep 15, 2014 10:20 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Creating a function given its vertex
Replies: 8
Views: 1401

Re: Creating a function given its vertex

PsylentKnight wrote:My professor ended up working this one out in class...

...a = -16/84 = -4/21

Okay, so I made a typo somewhere. I used the same method but got a = -4/13. Oops! :oops:
:clap:
by maggiemagnet
Fri Sep 12, 2014 4:00 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Creating a function given its vertex
Replies: 8
Views: 1401

Re: Creating a function given its vertex

How about if we set a = 1 and hope we can find something that works? a= 1 = (4x+1)/(x-3)^2 (x - 3)^2 = 4x + 1 x^2 - 6x + 9 = 4x + 1 x^2 - 10x + 8 = 0 But if you do the Quadratic Formula you get 2 solutions and you have to get only 1 because of what tangents do. So maybe try timesing it out: a(x^2 - ...
by maggiemagnet
Fri Sep 12, 2014 11:20 am
Forum: Advanced Algebra ("pre-calculus")
Topic: Creating a function given its vertex
Replies: 8
Views: 1401

Re: Creating a function given its vertex

...the tangent would cross the parabola at two points, correct? Or at least it appears that way in the links you posted. No, the crossing-at-two-points lines were the "secants", not the tangents. I don't understand where you're getting the m = 2(X - h) from. From calculus which you don't ...

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