4x-12 y= 5 find slope and y-intercept?
4x-12y=5 The first thing you do is try to get the equation into y=mx+b form. M would be the slope and b would be the y-intercept. So subtract 4x from both sides. -12y= -4x+5 Divide by -12, in order to isolate y. y= 1/3x -5/12 Now the equation is in y=mx+b form. Now just find the slope (m) and y-intercept (b) by looking at the equation. Slope: 1/3 Y-intercept: -5/12 =]
Parallel to the graph of 3x + 2y = 5; y-intercept (0, −1)?
the slope of 3x - 2y = 7 might want to nicely be stumbled on utilising m = -A/B the position A is the co-useful of x and B is the co-useful of y. so your slope is -3/-2 or basically 3/2. if the line you want to discover is parallel, meaning they share a similar slope. next you want the line to go via (-3,-5) so plug in -3 for x and -5 for y. 3(-3) - 2(-5) = C -9 + 10 = a million so 3x - 2y = a million will go via (-3, -5) and performance a slope of three/2 or in slope-intercept kind, y = 3/2x - a million/2
Find the slope and y-intercept - 3y - 2x = 5 + 9y - 2x?
I can figure most of these kind out if it is only 1 set of numbers, but I have never done them this way. I have searched for examples but have not found any like this. I work on my homework everyday for at least 6 hours and am still failing. This is my last class and I don't think I am going to make it to graduate. Thank you so much for any help. It is GREATLY appreciated !!!
Put the given eq. in the slope form(y=mx+c).5x+2y+7=02y= -5x -7y = (-5/2).x+(-7/2). by compairing with y=mx+c.m = -5/2 . Slope = -5/2 , Answer.Now put the eq. in the intersepted form( x/a+y/b=1).5x+2y+7=0or 5x+2y=-7Divide both side by -75x/-7+2y/-7= -7/-7x/(-7/5)+ y/(-7/2)= 1 . by compairing withx/a+y/b= 1a= -7/5 units , b= -7/2 units.x- intersept = -7/5 units.y- intersept= -7/2 units. Answer.
How to convert this to slope intercept form?
Slope intercept form looks like this: y = mx + b, where m is the slope of the line and b is the y-intercept. 5x + 12y - 12 = 24 5x + 12y = 36 <--- this is Standard Form 12y = -5x + 36 y = -5/12x + 3 <--- this is Slope-Intercept Form In accordance with the information I gave earlier, the slope is -5/12 and the y-intercept is 3. Written as a point (if your teacher requires that) it is (0, 3). Perpendicular lines have slopes that are negative reciprocals of each other. So, the slope of the other line would have to be 12/5. Parallel lines have the same slope. So, the slope would be -5/12.
10 points! whats the slope-intercept form of x+2y=5?!?
this is actually a straight line (y=mx+c) where c is the y-intercept and m is the gradient. if m is positive, the line is going up like this /. if m is negative, the line is going down. x+2y=5 2y=5-x (bring the x over. when a positive x is brought over, it becomes a negative. when a negative x is brought over, x becomes positive.) y=5/2-1/2x (divide both sides by 2 as according to the formula of the straight line which is written above) since the number before x is a negative number, the line is going down. whils 5/2 is the y-intercept whereby the line that cuts at the y-axis.
Write the linear equation 5x – 2y = –8 in slope-intercept form.?
number 4 is the answer! you have to write the equation in the y=mx+b format so -2y=-5x+(-8) then you divide everything by -2 so then you get y= -5/-2x+(-8)/-2 which you can reduce to y=5/2x+4
Question doesn’t make much sense as asked.I’ll assume you meant something like this:What is the standard equation OF A CIRCLE with centre at the origin AND tangent to 3x-2y = 5?Equation of circle : [math]\quad (x-h)^2 + (y-k)^2 = r^2 \quad[/math] where:[math](h, k) =[/math] centre of circle [math]= (0,0)[/math][math]r =[/math] radius of circle [math]=[/math] distance from origin [math](0,0)[/math] to line [math](3x-2y-5=0)[/math] [math]r = \dfrac{\left|3(0)-2(0)-5\right|}{\sqrt{3^2+2^2}} = \dfrac{5}{\sqrt{13}}[/math]Equation of circle:[math]\boxed{\boldsymbol{x^2 + y^2 = \frac{25}{13}}}[/math]Desmos graph
Find the lines tangent to the circle x^2+y^2-2x+4y+1=0 and parallel to the line 5x-12y=7?
The slope of the tangent is equal to the slope of the given line as they are parallel. The slope of given line is obtained by arranging it in standard form. 5x - 12y = 7 =>12y = 5x - 7 => y = (5/12)x - 7/12 the slope of tangent and line = 5/12 next find out the center and radius of the givencircle arrnaging circle in standard form x^2 - 2x + y^2 + 4y = -1 complete the squares x^2 - 2x +1 + y^2 + 4y + 4 = -1 + 1 + 4 (x - 1)^2 + (y + 2)^2 = 4 so center is (1,-2) and radius = sqrt(4) = 2 let the equation of tangent is y1 = mx1 + b, where m is slope and b= y-intercept substituting slope,m = 5/12, the equation of tangent is y1 = (5/12)x1 + b 12y1 = 5x1 + 12b 5x1 - 12y1 + 12b = 0 ----eqn (1) The perpendicular distance from a point (m,n) to the line ax + by + c = 0 is given by d = (am +bn + c)/sqrt(a^2+b^2) since radius and tangent are perpendicular, the distance from center to (1,-2) to the tangent line 5x1 - 12y1 + 12b is equal to radius so [5(1) -12(-2) + 12b]/(sqrt(169) = 2 5 +24 + 12b = 26 12b = -3 b = -3/12 = -1/4 substituting b value in 5x1 - 12y1 + 12b = 0 5x1 - 12y1 + 12(-1/4) = 0 5x1 - 12y1 - 3 = 0 So the equation of tangent to the given circle is 12y = 5x - 3
Well, for starters, when you’re talking about perpendicular and parallel slopes, you usually want to put the equation into point-slope (y=mx+b) form, as follows:[math]-8x-2y=-5[/math][math]2y=-8x+5[/math][math]y=-4x+(5/2)[/math]From this we see that your starting line has a slope of -4, therefore, any equation of the form [math] y=-4x+b[/math] will be parallel.Perpendicular lines have a slope that is equal to the negative reciprocal of the slope, in this case, 1/4. So, any equation of the form [math]y=(1/4)x+b[/math] will be perpendicular to your line