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Algebra 2 homework help?

Algebra 2 homework help? Topic: Homework help algebra slopes
June 16, 2019 / By Brandt
Question: So my teacher didn`t teach us how to find the slope of fractions and I have about 4 problems i need help on. Thanks in advance for the help!!! 2/3x+y/3=-1/3 (need in y=mx+b format) A/Dx+B/Dy=C/D (3/2,-1/2) (-2/3,1/3) - slope of line (0,-1/2) (7/5,10)
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Adolph Adolph | 10 days ago
y=mx+b is slope-intercept form, with the (x,y) being a point on a graph, the m is the slope and the b is the y-intercept. use order of operations to put y by itself to find the slope. 1) 2/3x + y/3 = -1/3 subtract 2/3x y/3 = -2/3x-1/3 multiply both sides by 3 y = -2x-1 so the slope is -2 2) A/Dx + B/Dy = C/D subtract A/Dx B/Dy = -A/Dx + C/D multiply both sides by D/B y = -ABx + CB slope is -AB to find the slope of two points, you use the formula (y2 - y1) / (x2 - x1) plugging in your numbers, this would look like: 3) (1/3+1/2) / (-2/3-3/2) do parenthesis first 5/6 / -13/6 to divide fractions, you multiple them by the reciprocal 5/6 x -6/13 = -5/13 and that is your slope.
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Adolph Originally Answered: Help with Algebra II homework?
Primo: 49+0.2m Ultimo: 16+1m U=4*P (Four times Primo charge) 16+1m = 4*(49+0.2m) 16+1m = 196+0.8m 1m-0.8m =196-16 0.2m =180 m =180/0.2 m = 900 miles (to the Ultimo charge be four times Primo charge).
Adolph Originally Answered: Help with Algebra II homework?
I'll try to help. I took Algebra II last year and am now in College Algebra. Primo price: 49+.2x=p *price* Ultimo price: 16+x=p What I did with these problems is guess and check. I guessed what I thought it would be then checked if I was right and if I was wrong I would adjust x accordingly. Sorry I couldn't help you work it out but there is my method
Adolph Originally Answered: Help with Algebra II homework?
Let it be x miles 4(49 +0.20x) = 16 + x 196 +0.80x = 16 + x 0.20x = 180 x = 900 miles ANSWER

Stevie Stevie
After looking to remedy drawback after drawback on yahooanswers, I truthfully feel so much folks dont be trained plenty if I supply them the solutions. Heres a VERY an identical drawback. Just fill within the values out of your possess drawback. (Hint: you're going to ought to use the Quadratic Formula on the finish!) Goodluck!!! QUESTION: The hypotenuse of a proper triangle has a period of thirteen cm. The sum of the lengths of the 2 legs is 17 cm. Find the lengths of the legs. Let a = period of 1 leg Let b = period of different leg We are informed that: a + b = 10.23 --> b = 10.23 - b [a million] The hypotenuse is eight.312. From Pythagorus: a² + b² = eight.312² [two] Substitute [a million] into [two]: a² + (eight.312 - b)² = 169 This simplifies to: 2a² - 34a + a hundred and twenty = zero Divide via two: a² - 17a + 60 = zero which elements: (a - five)(a - 12) = zero and has roots: a = five, 12 Substitute into [a million] and get: b = 12, five The lengths of the legs are five and 12 cm.
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Stevie Originally Answered: Algebra 2 Homework Help??
each time you make a new triangle you are basically putting a new longer row beneath the blocks you already have set up and the new row is only 1 block longer than the last so first triangle = 1 second = 1 + 2 = 3 blocks third = 1 +2 +3 = 6 blocks fourth = 1 + 2 + 3 + 4 = 10 blocks fifth = 1 +2 +3 +4 +5 = 15 blocks six = 1 +2 +3 +4 +5 +6 = 21 blocks ... going by this same pattern, ninth should be 1 +2 +3 +4 +5 +6 +7 +8 +9 = 45 blocks BTW there is a formula for this that Gauss figured out. n(n+1)/2
Stevie Originally Answered: Algebra 2 Homework Help??
The ninth triangle will have 9 blocks at the base, 8 at the second level, 7 at the third, etc., and there will be nine levels in total. So we want the sum of the numbers from 1 to 9, which is: (1 + 9) / 2 * 9 = 10 / 2 * 9 = 5 * 9 = 45 Therefore, 45 blocks are needed for the 9th triangle.

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