Algebra II problem-

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Math... really not my strong suit.

This is more like Algebra I, but my class is Algebra II, so whatever.
Directions: Write an equation in slope-intercept form for the line that satisfies each set of conditions.

Problem: slope 3/2, passes through (-5,1)

wtf do I do. So far I have...

y = 3/2 (-5) + b

1 = -15/2 + b

is that the answer?
 
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Having the ESF Forum do your math homework? Oh well, I'm bored...

They want an equation in slope-intercept form, a common form for 2D graphing. It's y = mx + b, where m is the slope, and b is the y-intercept. By pluging the point they give you into the equation, you solve for the intercept that shifts the line to include that point. You half-solved for b. Just find it, and write an equation that uses it and the slope given.
 
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b = 1+ 15/2 = 17/2

Therefore, the equation will go y=3/2*x + 17/2 <-- this is slope-intercept form. Dont comfuse it with standard form, which is Ax + By + C = 0, or something like that.

Maybe you want to do my homework for me... more specifically, question 4d. Everyone I know in MATH 100 is freaking out over this assignment.
 
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frsrblch said:
b = 1+ 15/2 = 17/2

Therefore, the equation will go y=3/2*x + 17/2 <-- this is slope-intercept form. Dont comfuse it with standard form, which is Ax + By + C = 0, or something like that.
I feel soo dumb... How do you do it that quick?
 
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since growler's question is solved anyone mind answering this one for me..

two times one number exceeds five times another number by 10 (form it into an algebraic equation)
 
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2 x X > 5 x (Y x 10)?

X = 'one number'
Y = 'another number'
 
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DarkSniper said:
two times one number exceeds five times another number by 10 (form it into an algebraic equation)
2x = 5y + 10
Take a look at it so you understand how/why it works.

I should post my next calculus assignment here. =o
 
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<--- took calc 3...i don't think you want to see my equations
 

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Who's gonna help me do my calculus :(
 
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Threads like this aren't allowed. You're not going to learn jack if other people solve it for you. If you need help, consult your classmates or talk to the teacher. Everyone should be more than willing to help.
 

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