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What is 1 divided by x divided by 2 divided by y?
0.5
X-y equals 5 and x squared-y squared equals 5 How do you work this out using simultaneous equations?
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How would you graph x 2?
y = X^2 Is this what you mean? functions are y values when x = 1; y =1 when x = -1; y = 1 when x = 2; y = 4 when x = -2; y = 4 when x = 3; y = 9 when x = -3; y = 9 get it? A upward opening parabola with its vertex at the origin.
How do you differentiate y equals x minus a over x?
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
What is the equation for a circle centered at -1 -2 and has a radius of 8?
(x + 1)2 + (y + 2)2 = 82(x + 1)2 + (y + 2)2 = 82(x + 1)2 + (y + 2)2 = 82(x + 1)2 + (y + 2)2 = 82
What is the derivative of y equals square root of x when x is positive?
y = square root of x y = x(1/2) y' = (1/2)x(1/2 - 1) y' = (1/2)x(-1/2) y' = (1/2)(square root of x)
What is the range of y equals negative 2x squared plus x?
y = -2x² + x y = -2 [ x² - (x/2) ] y = -2 [ x² - (x/2) + (1/4) ] + (1/2) y = (1/2) - 2[ x - (1/2) ]² ≤ (1/2) Hence, the Range of y is : -∞ < y ≤ (1/2). It can be written as the interval ( -∞, 1/2 ]. ........ Ans. _____________________________________ Happy To Help ! _____________________________________
What is the second derivative of xsqrtx-1?
let y= xsqrt(x) -1 y= x^(3/2) -1 ---- since xsqrt(x) is the same as x^(3/2) y' = (3/2) x^(3/2-1) y' = (3/2) x^(1/2) y'' = (3/2) (1/2) x^(1/2-1) y'' =(3/2)(1/2) x^(-1/2) y'' = 3/4x^(1/2) y'' = 3 / 4sqrt(x)
Differentiate 3 Sq root x?
y = 3√x y = 3x^(1/2) y' = 3(1/2)x^(1/2 -1) y'= (3/2)x^(-1/2) y' = 3/[2x^(1/2)] y' = 3/(2√x)
Y divided 2 plus x use x equals -1 and y equals 2?
Y / (2+x)If x = -1 and y=2, thenY / (2+x) = 2/(2-1) = 2/1 = 2
What is 1 divided by x divided by 2 divided by y?
0.5
What are y equals x plus 2 y equals -x plus 4?
y = x + 2 y = -x + 4 x + 2 = -x + 4 2x + 2 = 4 2x = 2 x = 1 y = x + 2 y = 1 + 2 y = 3 (1, 3)
How do you substitute x plus y equals 5 and x equals y plus 1?
Eq 1: + y= 5 Eq 2: x = y + 1 substitute y + 1 for x in equqtion 1 y+1 + y = 5 2y + 1 = 5 2 y = 4 y = 2 substitute 2 for y in eq 2 x = 2 + 1 x = 3
Does the graph x-y2 equals 1 represent x as a function of y?
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
How do you graph x-y equals 2?
x-y=2 -y=(2-x) y = -2 + x make t chart and you get x:1, 2 , 3, 4, 5 Y:-1, 0, 1, 2, 3
Find the function whose inverse is the same as the function itself?
f ( x ) = (x-2)/(x-1)if y = (x-2)/(x-1)yx-y= x - 2yx-x= -2+yx(y-1)=y-2x = (y-2)/(y-1)so g ( x ) the inverse function is also (x-2)/(x-1)
X-y equals 5 and x squared-y squared equals 5 How do you work this out using simultaneous equations?
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