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Finding the Area of a Triangle  TOPIC_SOLVED

http://img52.imageshack.us/img52/9089/geometrytrianglequestio.png Thank you for your time, I just need some guidance/hints on this. It is a right angled triangle with the height being 3.7 cm and a length of the base being 3.5 cm. I tried to construct another triangle on top of the 3.5 cm length in ...
by cruxxfay
on Tue Aug 14, 2012 11:00 am
 
Forum: Geometry
Topic: Finding the Area of a Triangle
Replies: 4
Views: 3369

Re: Finding the Area of a Triangle  TOPIC_SOLVED

The question is: "Find the areas of these shapes. Give your answer to 3 significant figures." There are 6 shapes in that exercise and this particular one is question 5. Graphically I am given 3 information, as you have already noticed. I am starting to wonder if they are missing one piece ...
by cruxxfay
on Wed Aug 15, 2012 3:25 am
 
Forum: Geometry
Topic: Finding the Area of a Triangle
Replies: 4
Views: 3369

Re: Finding the Area of a Triangle  TOPIC_SOLVED

Thank you for the answer. I was doing a Pre-IB course and thought there would be an advanced way of solving this. But anyway thanks :D
by cruxxfay
on Thu Aug 16, 2012 3:04 am
 
Forum: Geometry
Topic: Finding the Area of a Triangle
Replies: 4
Views: 3369

Polynomial Functions Questions  TOPIC_SOLVED

Hello, Thanks for taking your time reading. I have encountered 2 problems as I was going through my textbook, so may I please kindly ask of some assistance on this. "Find the polynomial f such that f(2x-1) = 16x^4 - 32x^3 +12x^2" And "When polynomial f is divided by (x-3) the remainde...
by cruxxfay
on Thu Sep 13, 2012 11:39 am
 
Forum: Intermediate Algebra
Topic: Polynomial Functions Questions
Replies: 3
Views: 3441

Re: Polynomial Functions Questions  TOPIC_SOLVED

So r(-1) = -4. Since we are dividing by a quadratic, then logically the remainder can have a degree of 0 or 1. Since r(x) is clearly not constant, then r(x) must be linear. Then r(x) = ax + b for some values a and b. One may then find the equation of the remainder as being r(x) = (3/2)x - (5/2). I k...
by cruxxfay
on Sun Sep 16, 2012 10:17 am
 
Forum: Intermediate Algebra
Topic: Polynomial Functions Questions
Replies: 3
Views: 3441

Logarithm and infinite geometric series  TOPIC_SOLVED

"Consider the infinite geometric series 1 + (2x/3) + (2x/3)^2 + (2x/3) ^3 +... For what values of x does the series converge?" I know convergence means the sum tends to a finite series, so I have used the equation Sn = u/(1-r) where u = first term and r = ratio Sn = 1/(1-2x/3) But then I h...
by cruxxfay
on Sat Nov 24, 2012 2:03 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Logarithm and infinite geometric series
Replies: 5
Views: 2183

Re: Logarithm and infinite geometric series  TOPIC_SOLVED

No, that is the entire question.
by cruxxfay
on Sun Nov 25, 2012 12:31 pm
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Logarithm and infinite geometric series
Replies: 5
Views: 2183

Re: Logarithm and infinite geometric series  TOPIC_SOLVED

I realized I could use the ratio test to solve that sequence, although I have yet to learn it at school. But then I am still confused with the a + log (base5) b part In fact this is what I got: 5^x = 3/5 x = log(base5)^3/5 = log(base5)^3 - log(base5)^5 = -1 + log(base5)^3 so a = -1 and b = 3? I don'...
by cruxxfay
on Mon Nov 26, 2012 5:04 am
 
Forum: Advanced Algebra ("pre-calculus")
Topic: Logarithm and infinite geometric series
Replies: 5
Views: 2183

Finding the phase shift  TOPIC_SOLVED

I understand that in asin(bx + c) + d, where c represents the phase shift. But then in a question like: f(x) = 7sin[6(x + pi/12)] + 3 the phase shift here is pi/12 (I thought it was pi/2) and f(x) = -3sin(2x + pi/2) - 5 the phase shift here is pi/4...again I thought it was pi/2. So how exactly do yo...
by cruxxfay
on Wed Jan 02, 2013 5:25 am
 
Forum: Trigonometry
Topic: Finding the phase shift
Replies: 1
Views: 1891

Graphing a trig function  TOPIC_SOLVED

The question is as follow: f(x) = (2+3sinx)/(4+3cosx) I know that the first thing you do is to try simplifying this down to 1 trig ratio, but well, I use cos(x) = sqrt[1-sin^2(x)], which got me nowhere. I have looked at the answer and the shape appears pretty weird to me so I have no idea where to s...
by cruxxfay
on Sat Jan 12, 2013 12:49 pm
 
Forum: Trigonometry
Topic: Graphing a trig function
Replies: 4
Views: 2999
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