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How do you Write an equation in the form of y equals kx for the following data?
There are no "following" data!
What are the values of k when the line kx plus y equals 4 touches the curve y equals x squared plus 8 on the Cartesian plane showing work?
If: y = x^2 +8 and kx +y = 4 or y = 4 -kx Then: x^2 +8 = 4 -kx So: x^2 +8 -4 +kx = 0 => x^2 +4 +kx = 0 Using the discriminant b^2 -4ac = 0: k^2 -4*1*4 = 0 => k^2 = 16 Therefore the values of k are: -4 or 4 Hence: y-4x = 4 and y+4x = 4 are tangents to the curve y = x^2 +8
What is the value of k when the line y equals kx plus 1.25 is a tangent to the curve y squared equals 10x?
Equations: y = kx +1.25 and y^2 = 10x If: y = kx +1.25 then y^2 = (kx +1.25)^2 =>(kx)^2 +2.5kx +1.5625 So: (kx)^2 +2.5kx +1.5625 = 10x Transposing terms: (kx)^2 +2.5kx +1.5625 -10x = 0 Using the discriminant formula: (2.5k -10)^2 -4(1.5625*k^2) Multiplying out the brackets: 6.25k^2 -50k +100 -6.25^2 = 0 Collecting like terms: -50k +100 = 0 Solving the above equation: k = 2 Therefore the value of k is: 2
What is the value of k in the line of y equals kx plus 1 and is tangent to the curve of y squared equals 8x?
y = kx + 1 y^(2) = 8x Square up the linear eq'n and substitute. Hence y^(2) = ( kx + 1)^(2) (kx + 1)^(2) = 8x Square out the brackets (kx)^(2) + 2kx + 1 = 8x (kx)^(2) + (2k-8)x + 1 = 0 Apply the discriminant b^(2) -4ac = 0 (2k - 8)^(2) - 4(k)^(2)(1) 4k^(2) - 32k + 64 - 4k^(2) = 0 32k - 64 = 0 32k = 64 k = 2 =64
What are the values of k when the line of y equals kx -2 is a tangent to the curve of y equals x squared -8x plus 7?
If: y = kx -2 and y = x^2 -8x+7 Then the values of k work out as -2 and -14 Note that the line makes contact with the curve in a positive direction or a negative direction depending on what value is used for k.
Find a equation of variation where y varies directly as x and y equals 0.8 when x equals 0.4?
direct variation: y = kx y = kx k = y/x = 0.8/0.4 = 2
A type of variation in the form y equals kx?
Y varies in direct proportion to x.
If y varies directly as x squared and y equals 150 when x equals 5 find y when x equals 4?
First you need to know that the equation you are looking for is y = kx^2. Then you need to substitute the numbers in: y = kx^2 150 = k5^2 150 = k25 6 = k Now that you know k, resubstitute it for the new value of y when x = 4: y = kx^2 y = (6)(4)^2 y = (6)(16) y = 96
How do you Write an equation in the form of y equals kx for the following data?
There are no "following" data!
The variables x and y vary directly when x equals -5 and y equals 21 what is the constant of variation?
y = kx k = y/x = 21/-5 = - 21/5
Suppose y varies directly with x and y equals 12 when x equals 6write an equation realting x and y?
y = kx ie y/x = k, k = 2, so y = 2x
What are the values of k when the line kx plus y equals 4 touches the curve y equals x squared plus 8 on the Cartesian plane showing work?
If: y = x^2 +8 and kx +y = 4 or y = 4 -kx Then: x^2 +8 = 4 -kx So: x^2 +8 -4 +kx = 0 => x^2 +4 +kx = 0 Using the discriminant b^2 -4ac = 0: k^2 -4*1*4 = 0 => k^2 = 16 Therefore the values of k are: -4 or 4 Hence: y-4x = 4 and y+4x = 4 are tangents to the curve y = x^2 +8
What is the value of k when the line y equals kx plus 1.25 is a tangent to the curve y squared equals 10x?
Equations: y = kx +1.25 and y^2 = 10x If: y = kx +1.25 then y^2 = (kx +1.25)^2 =>(kx)^2 +2.5kx +1.5625 So: (kx)^2 +2.5kx +1.5625 = 10x Transposing terms: (kx)^2 +2.5kx +1.5625 -10x = 0 Using the discriminant formula: (2.5k -10)^2 -4(1.5625*k^2) Multiplying out the brackets: 6.25k^2 -50k +100 -6.25^2 = 0 Collecting like terms: -50k +100 = 0 Solving the above equation: k = 2 Therefore the value of k is: 2
If y Varies Directly as x to the 2nd power and y equals 198 when x equals 6 find y when x equals 2?
y=kx^2 hence k=198/36. now y=198/36*(2)^2 y=22
If y varies directly as x and y equals 18 when x equals 3?
y varies directly as x y=kx, where k is a constant y=18 when x=3 18=3k k=6 y=6x
Find the variation constant and an equation of variation where y varies directly as x and y equals 10 when x equals 37?
y = kx: 10 = 37k so k = 10/37 and y = 10x/37
Y varies directly as x find the constant of variation k if y equals negative 5 x?
Y=kx+m Y=-5x+m K=-5
