Help me with High School Geometry

[Smartly Pretty 0]

The measure of the supplement of an angle is 40 less than 3 times its complement. Find the measure of the angle, its supplement, and its complement.

 

I'm really awful at math.

[FroGG_NeaL 0]

I guess what it all comes down to is that the angle of my dangle is inversely proportional to the heat of my meat.

[MFer 0]

Angle=25

Complementary=65

Supplementary=155

 

Don't ask.

[Just John 0]

Edit: MFer got it.

[CanadianChris 0]

To help, supplement = 180 - angle, and complement = 90 - angle.

 

Aside from that, it's an algebra problem.

[Smartly Pretty 0]

2.04] Write the equation of the line that passes through the points (-5, 1) and (-5,-2). Show all your work for credit.

[Smartly Pretty 0]

That problem's fucking bullshit because that's clearly algebra

 

and yeah, I'm gonna bump this thread a lot this summer

[CanadianChris 0]

There's no work to show. It's a vertical line, so it has infinite slope and no y-intercept. The equation is x=-5.

[Smartly Pretty 0]

CC, you're gonna be my go to guy this summer.

[CanadianGuitarist 0]

CC, you're gonna be my go to guy this summer.

I'd wait until the hockey draft is over until he becomes your "Go To" guy.

[TheSSNintendo 0]

I hated Geometry in high school. Failed it in 10th grade, but managed to pass the SOL. Took it over again in 11th grade, but felt that I wasn't doing well enough (had a D), so I dropped it and got sent to Algebra 1B/Geometry A. And then I had to do Geometry B my senior year, and managed to do better.

[Red Baron 0]

http://forums.thesmartmarks.com/index.php?showtopic=91248

[DrVenkman PhD 0]

You know bps, when I first saw that thread I thought it was about BEDMAS not being taught. I remember when I was growing up teachers brought up that older folk had never heard of the order of operations, thus McDonald's skill testing questions were always incorrect.

[Guest]

I also failed Geometry in 10th grade, but only because I didn't want to do any homework. Homework never was my thing.

 

I went to Summer School and got an A. Easy class, but I don't remember any of it.

[At Home 0]

I've got an A in every math class I've ever taken... shit just comes kinda easy for me. But it's been a while since I've taken a math class though, I finished all my credits early to not take a class second semester senior year.

[CanadianChris 0]

I have a degree in statistics, so I can say I usually did pretty well in math.

[AboveAverage484 0]

I'm pretty efficient at working with numbers and algebra in real world settings, but I could never quite get it down in school. I worked hard to get B's in high school Algebra and made a C- in Geometry.

[MFer 0]

I found geometry and trigonometry to be fairly easy in high school. I had a little more trouble with algebra but I did well enough.

 

EDIT: I also took calculus in high school. Pretty rough stuff, but I got my C and moved on. In college, I basically just took the standard algebra and pre-calculus courses to get that requirement out of the way.

[Boon 0]

I got as high as calculus in high school. Failed that, went to school for political science, haven't had to use anything higher than eighth grade algebra since.

[Smartly Pretty 0]

Part a. Write the equation of the line that is parallel to the line 2x-3y= 0 and contains the point (6,1).

Part b. Write the equation of a line that is perpendicular to the line 3x-y= -2 and contains the point (1,3).

[Guest !!!]

You know bps, when I first saw that thread I thought it was about BEDMAS not being taught.

It's PEMDAS where I come from.

[CanadianChris 0]

a) Since the line is parallel to 2x-3y=0, it has the same slope. Rewrite the equation as 3y=2x => y=2/3x. So, slope is 2/3, and the line is y=2/3x + b.

Now, the line passes through (6,1). To find the y-intercept, plug (6,1) into the equation y=2/3x + b:

 

y=2/3x + b

(1)=2/3(6) + b

b=1-4

b=-3.

 

This makes the equation of the line y=2/3x - 3.

 

Follow the same steps for b).

[Smartly Pretty 0]

I'm glad you did that all out, because this isn't like summer school or anything, it's online school, so I had absolutely no idea how to fucking do that, but it's been on the past, like, 8 assignments. Thanks man.

[Smartly Pretty 0]

Given: Triangle ABC with mangleA = 2x2-2x+44, mangleB = 5x2+30x+50 and mangleC = x2+20x+30.

Find each angle measure and classify triangle ABC as acute, right or obtuse.

[At Home 0]

mangle. Hah.

[Just John 0]

Given: Triangle ABC with mangleA = 2x2-2x+44, mangleB = 5x2+30x+50 and mangleC = x2+20x+30.

Find each angle measure and classify triangle ABC as acute, right or obtuse.

 

Are those 2's are supposed to be exponents? Just for future reference, to denote that in typing you can write it as 2x^2, etc. Anyway, your problem...

 

A triangle as 180 degrees, so A + B + C = 180, which gives you this mess:

 

(2x^2 - 2x + 44) + (5x^2 + 30x + 50) + (x^2 + 20x + 30) = 180

 

Just kind of simplify it and work from there. Looks like it'll be a decimal answer.

 

 

[CanadianChris 0]

It's actually simpler than that. If you add that big-ass equation together, you get 8x^2 + 48x - 56 = 0. Divide out the 8 and you're left with x^2 + 6x - 7 = 0. Now you just solve it as a quadratic equation:

 

x^2 + 6x - 7 = 0

(x+7)(x-1) = 0

x=-7 or x=1

 

Now, plug x=-7 into the three angle measurements, and then plug x=1 into them. One works (all the angles are greater than zero and add to 180), one doesn't.

 

As for the type of triangle, an acute triangle has three acute angles (<90 degrees), a right triangle has one right angle and two acute angles, and an obtuse triangle has one obtuse angle (>90 degrees) and two acute angles.

[Dr. Zoidberg 0]

....I am so fucking lost...

[Broward83 0]

It's actually simpler than that. If you add that big-ass equation together, you get 8x^2 + 48x - 56 = 0. Divide out the 8 and you're left with x^2 + 6x - 7 = 0. Now you just solve it as a quadratic equation:

 

x^2 + 6x - 7 = 0

(x+7)(x-1) = 0

x=-7 or x=1

 

Now, plug x=-7 into the three angle measurements, and then plug x=1 into them. One works (all the angles are greater than zero and add to 180), one doesn't.

 

As for the type of triangle, an acute triangle has three acute angles (<90 degrees), a right triangle has one right angle and two acute angles, and an obtuse triangle has one obtuse angle (>90 degrees) and two acute angles.

 

...and this is why I'm a History major.