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From the equation X-Y equals 2 what is Y when X equals 3?
X - Y = 2 When X = 3 you have 3 - Y = 2 Moving Y to RHS: 3 = Y + 2 Moving 2 to LHS: 1 = Y
What is inverse of y equals 2radical x plus 3?
To find the inverse you switch the x and the y and then solve for y. x=2 radical( y + 3) radical(y + 3) = x/2 y+3= (x/2)² y = (x/2)² -3 So the answer is y = (x/2)² -3
What is y equals 2- -3?
5
What is the LCM of 6y3 and 18y4?
Prime factor both numbers or expressions and use the factors the number of times it was used the MOST in either number. 6y^3=2*3*y*y*y 18y^4=2*3*3*y*y*y*y 2*3*3*y*y*y*y=18y^4
How to factorise I x power 6 - y power 6 2 x power 12- y power 12?
Explanation: The difference of squares identity can be written: a 2 − b 2 = ( a − b ) ( a b ) The difference of cubes identity can be written: a 3 − b 3 = ( a − b ) ( a 2 a b b 2 ) The sum of cubes identity can be written: a 3 b 3 = ( a b ) ( a 2 − a b b 2 ) So: x 6 − y 6 = ( x 3 ) 2 − ( y 3 ) 2 = ( x 3 − y 3 ) ( x 3 y 3 ) = ( x − y ) ( x 2 x y y 2 ) ( x y ) ( x 2 − x y y 2 ) If we allow Complex coefficients, then this reduces into linear factors: = ( x − y ) ( x − ω y ) ( x − ω 2 y ) ( x y ) ( x ω y ) ( x ω 2 y ) where ω = − 1 2 √ 3 2 i = cos ( 2 π 3 ) sin ( 2 π 3 ) i is the primitive Complex cube root of 1 .
What is y (2 A) 3?
what is y+(2+a)+3
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)
What is y squared multiplied by y cubed?
y^2 X y^3 = y^(2 + 3) = y^5 You can only do this if the coefficient 'y' is the same for both terms. Remember y^2 = y X y y^3 = y X y X y Hence y^2 X y^3 = y X y X y X y X y = y^5 Similarly for division/subtraction y^3 / y^2 = y^(3 - 2 ) = y^1 = y The power of '1' is trivial and not normally shown. NB You CANNOT do z^2 X y^3 by adding the indices. z^2 X y^3 is (z^2)*(y^3)
Find the vertex of y equals 3x2-4x equals 3?
y=3x^2-4x=3 y=3x^2-4x-3 y = ax^2 + bx + c is the parabola y = 3x^2-4x-3 x-coordinate of vertex is -b/2a = -(-4) / (2)(3) = +4 / (2)(3) = 4/6=2/3 y coordinate of the vertex: y = ax^2+bx+c y = 3(2/3)^2-4(2/3)-3 y= 3(4/9) -8/3 -3 y=12/9-8/3-3 = -13/3 Vertex : ( 2/3, , -13/3 )
How do you answer something like y 3x negative 2?
If you mean y=3x-2, you can solve for x as follows: 3x-2+2 = y+2, x = (y+2)/3 = y/3 + 2/3
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)
Y2 - 4 over y plus 3 equals 2 - y - 2 over y plus 3?
(2y-4)/(y+3)=(2-y-2)/(y+3) --> multiply both sides by (y+3) 2y-4=2-y-2 --> 2-2=0 (right hand side) 2y-4=-y --> add y to both sides 3y-4=0 --> add 4 to both sides 3y=4 --> divide both sides by 3 y=4/3
How to factorise I x power 6 - y power 6 2 x power 12- y power 12?
Explanation: The difference of squares identity can be written: a 2 − b 2 = ( a − b ) ( a b ) The difference of cubes identity can be written: a 3 − b 3 = ( a − b ) ( a 2 a b b 2 ) The sum of cubes identity can be written: a 3 b 3 = ( a b ) ( a 2 − a b b 2 ) So: x 6 − y 6 = ( x 3 ) 2 − ( y 3 ) 2 = ( x 3 − y 3 ) ( x 3 y 3 ) = ( x − y ) ( x 2 x y y 2 ) ( x y ) ( x 2 − x y y 2 ) If we allow Complex coefficients, then this reduces into linear factors: = ( x − y ) ( x − ω y ) ( x − ω 2 y ) ( x y ) ( x ω y ) ( x ω 2 y ) where ω = − 1 2 √ 3 2 i = cos ( 2 π 3 ) sin ( 2 π 3 ) i is the primitive Complex cube root of 1 .
What is the product of 7 and y to the third power when y 2?
To write this algebraically: 7(y^3)|y = 2 Substitute 2 for y: 7(2^3) 2^3 = 8, so substitute 8 for (2^3): 7*8 = 56
What point is the intersection of the graphs of the lines 2X-Y equals 3 and X plus Y equals 3?
2X-Y=3 X+Y=3 ---------- 3X = 6 X=2 2(2)-Y=3 4-Y=3 Y=1 Point of interesection: (2,1).
2x equals 3-y 2y equals 12-x?
2x=3-y 2y=12-x y=3-2x 2(3-2x)=12-x 6-4x=12-x -3x=6 x=-2 2(-2)=3-y -4=3-y y=7 x=-2 y=7
How do you factor y cubed plus y squared minus 6?
y^3 + y^2 - 6 (y+2) (y-3) i think
