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If you mean: x+y = 30 and the value of y is 15 then the value of x is also 15
The tangent to 2x3 - 3x2 - 8x + 9 at x = 2 is y = 4x - 11 The tangent to y = 2x3 - 3x2 - 8x + 9 at x = 2 has the same gradient as the curve at that point; to find the gradient, differentiate: dy/dx = 6x2 - 6x - 8 which at x = 2 is: gradient = 6 x 22 - 6 x 2 - 8 = 4 At x = 2, y = 2 x 23 - 3 x 22 - 8 x 2 + 9 = -3 The equation of a line through point (xo, yo) with gradient m is: y - yo = m(x - xo) Thus the equation of the tangent to the line at x = 2 is: y - -3 = 4(x - 2) ⇒ y = 4x - 11
Y = X - 16 X= Y + 16 X is the original value. Y is the new value of the original value - 16.
If y varies directly as x and y is 36 when x is 9, then y is always four times the value of x. So if y is 12, then x is 3.
You already know the value of x is 4. To solve the value of y = 64x, mutlply 64 x 4. Hence, y = 64*4 = 256.
If x = -7, then x3 = -343 so that 2x3 = -686 y = 2 then 15 y = 30 So 2x3 + 15y = -686 + 30 = -656
x = -7, y = 2 ⇒ 2x3 + 15y = 2 x (-7)3 + 15 x 2 = 2 x (-343) + 15 x 2 = -656
In expressions such as "x-y", both "x" and "y" can have any value. The value of "x-y" will depend on what the value of "x" and the value of "y" are.
If you mean: x+y = 30 and the value of y is 15 then the value of x is also 15
The value of x + y is indeterminate. You need the values of both x and y to determine x + y.
y is a function of x iffor each value of x (in the domain) there is a value of y, andfor each value of y (in the range) there is at most one value of x.
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
if 2x + y = 7, what is the value of y when x = 3?
y = 100 - 4x ( a ) Find the value of y when x = 20.
The tangent to 2x3 - 3x2 - 8x + 9 at x = 2 is y = 4x - 11 The tangent to y = 2x3 - 3x2 - 8x + 9 at x = 2 has the same gradient as the curve at that point; to find the gradient, differentiate: dy/dx = 6x2 - 6x - 8 which at x = 2 is: gradient = 6 x 22 - 6 x 2 - 8 = 4 At x = 2, y = 2 x 23 - 3 x 22 - 8 x 2 + 9 = -3 The equation of a line through point (xo, yo) with gradient m is: y - yo = m(x - xo) Thus the equation of the tangent to the line at x = 2 is: y - -3 = 4(x - 2) ⇒ y = 4x - 11
In most cases, x is independent and y is dependent. That is, you choose the value of x, but this x-value will decide the corresponding y-value.
If: y = x+b then y-x = b