Basic Calc Function  TOPIC_SOLVED

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Basic Calc Function

Postby Egibson on Mon Jan 18, 2010 3:22 am

I am in an elementary calculus class and I am struggling with my first assignment 1) because my new teacher is tough to understand. and 2) because my course load has increased significantly, so I have realized I need more time dedicated to my math.... anyways any help IS appreciated. :oops:


1) So I'm Sitting here looking at a function and I am getting ready to throw the book out the window. so the questions is what do all family members of f(x)= 1 + m(x + 3) have in common. so i have placed different values in the x, and then all i get is 1 + (insert growing # here)m like i know that i have my formulas y=mx+b, or y-y1=m(x-x1) but I just can't seem to find what they have in common.

2) This is my second question. I know it is a lot for a first post however i will try and provide as much as help when I'm smarter :D . anyways the question is... Show the area of a triangle with sides of lengths a and b and with including angle(theta) is
A=1/2absin(theta)
all i have is a right triangle in quadrant 2(thats sin right??) with a on my x axis, b on my y axis, (theta in the 45 degree angle) and I am lost.
Egibson
 
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Re: Basic Calc Function

Postby dicktahoe on Mon Jan 18, 2010 6:41 pm

See what you think about this for the first problem. Let x= -3. Then the point at which all linear equations cross for any value of m, is (-3, 1). The thing the family has in common is that no matter the value of m, all functions cross at the point (-3, 1). so (-3, 1) is a solution to this family of linear equations.
dicktahoe
 
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Re: Basic Calc Function  TOPIC_SOLVED

Postby dicktahoe on Mon Jan 18, 2010 7:31 pm

For the 2nd problem. Let a=hypotenuse of the right triangle, and b = the adjacent side, and theta is the angle between a and b. Then let the opposite side of the triangle, which is the height of the triangle, be h. Now sin (theta) = h/a, and h= a sin (theta). Since the Area of a triangle is 1/2 bh, Substituting, h= a sin (theta) into your equation A = 1/2b [asin(theta)] we get A = 1/2 a h. Done.
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