Write an equation in standard form?
The standard form is always written in the form: Ax + By = C where the coefficients (A,B,C) are "relatively prime" integers. so.... #1 We are given a slope and an intercept (a point on the line), so we write the equation in y = mx+b form using the point-slope form: y - k = m(x - h) where m is the slope, h is the x-coordinate, and k is the y-coordinate: y - (-4) = (2/3)(x-0) --> y + 4 = (2/3)x --> y = (2/3)x - 4 Since we have a fraction we need to get rid of it by using a common denominator: y = (2/3)x - (12/3) then multiply the entire equation by the common denominator (3) to get: 3y = 2x - 12 Now re-write the equation in Ax + By = C form: 2x - 3y = 12 #2 Do the same thing... the slope is 2 (see it next to the "x") the lines are parallel so they have the same slope. The point is once again given (1,3) #3. This time the slope is perpendicular to the given slope (2/5), so flip the fraction and make it negative to get m = -(5/2). The point is (3,0) #4. No slope means m=0 and the point is (2,4) HTH -- Be sure the recheck my basic math skills, but I know the concepts are correct. @Buffy S: You did a good job with this, but no-slope is not the same as infinite slope, so there is only one equation for #4. EDIT: "relatively prime" mean that you can't reduce the coefficients any further. For instance if you have (A=2, B=3, C=12) they are "relatively prime". But if you have (A=6, B=3, C=12) you would want to divide them all by 3 and get (A=2, B=1, C=4).
Write the standard form of the equation of each line??Help!!?
maybe you should have listen in class you just put the in y=mx+b form 1.) -2y-5x=10 -2y=5x+10 y=-5/2x - 5 2.)1=-1/3y+1/3x 1/3y=1/3x-1 y=x-3 3.)3y+9x+15=0 3y=-9x-15 y=-3x-5 4.)-2y+7=-x -2y=-x-7 y=1/2x + 7/2 5.)5y=-2x y=-2/5x 6.)24y-9x=84 24y=9x+84 y=3/8x+ 7/2 7.)y=-3-x y=-x-3
Write the equation y=1/4x-2 in standard form using only integers and a positive coefficient for x.?
All you do is make the equ. = 0 so move all to the figures to the left. When you get a negative, just multiply both sides by -1. Therefore, you get this... 1/4x - y - 2 =0 To get it into integers, get rid of the denominator multiply by 4 x - 4y - 8=0 Cheers
How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?
There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.
Give the equations below rewrite the equation in standard form?
Standard form: Ax + By = C A, B, and C must be whole numbers A must be positive. 1) 5x+y = 7 2) 2y = x - 18 -x + 2y = -18 x - 2y = 18 3) 4y = 3x + 40 -3x + 4y = 40 3x - 4y = -40
How many roots of equation ax^2+b|x|+c=0?
It may have max 4 rootsMod is present so root of opposite sign is also a rootRoot with pairs will exist so even no. Of roots only possibeSo 3 roots not possibleWhen zero will be a root then we would have zero as a repeated root hence total root will be 4 but when dintinct roots are asked then 3.In this case no. Of solution is 3.No. Of solution and roots are different keep that in mindSimilarly 2 root is also possible1 root not possible as we have even no. Of roots in this case as modulus presentNo root is also a caseCheck discriminant sign ,draw the graph of equation without taking modulus.don't draw the rough graph on the negative x axis.then reflect the graph on the positive side to the negative x axisThe resulting graph will be of the required equation .u can find many things about the equation from a graph
Help with math: standard form of a linear equation?
So, I'm not doing too good in math right now so I took the day off from school to study... and i need help. We just started this yesterday but i'm probably gonna be having a test on this chapter anywhere from wednesday-thursday so please help! I have to Write the equation in standard form with integer coefficients. 1. y=(-5x)+2 2. 3x+9=7/2y 3. y=(-3/4x)+5/4 and then i have to: write an equation in standard form of the line that passes through the given point and has the given slope 4. (-8,3) m=2 5. (-6, -7) m=(-1) 6. (2,9) m= (-7) Please don't just give me answers b/c i seriously need like an A or a B on the test and i need to know how to solve these. I've tried looking these up on the internet but none of them show me step by step instructions which is what i really need. Thanks :)
Algebra 2 math equation to standard form HELP :(?
Hi, I always require that standard form starts with a positive number. 1) m = -3, B = 5 y = mx + b is slope intercept form to start. y = -3x + 5 3x + y = 5 <==STANDARD FORM 2) m = -3/2, passes through (4, -7) y - y1 = m(x - x1) is point slope form with slope m through the point (x1,y1) y + 7 = -3/2(x - 4) y + 7 = -3/2x + 6 3/2x + y = -1 3x + 2y = -2 <==STANDARD FORM 3) passes through (-1, 3) and (-6, -7) m = (y2 - y1)/(x2 - x1) to find the slope m Then y - y1 = m(x - x1) is point slope form with slope m through the point (x1,y1) -7 - 3 --------- = 2 = m -6 + 1 y - 3 = 2(x + 1) y - 3 = 2x + 2 -2x + y = 5 2x - y = -5 <==STANDARD FORM I hope that helps!! :-)