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what is y+(2+a)+3
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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 )
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
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).
What is the y intercept and slope of y 3 over 2 x?
There is a missing equals. When the equation of a line is written in the form: y = mx + c m is the gradient/slope of the line; and c is the y intercept. If your line is ``y = 3/2 x'' then y = 3/2 x → y = 3/2 x + 0 → slope is 3/2 & y intercept is 0.
What is 3y divided by 2y?
1.5
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
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 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 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)
What is y equals 2- -3?
5
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)
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
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 .
Where are the points of intersection of the line x -y equals 2 with the curve x2 - 4y2 equals 5 by completing the square?
If: x -y = 2 then x = 2 +y If: x^2 -4y^2 = 5 then (2 +y)^2 -4y^2 -5 = 0 Removing brackets: 4 +4y +y^2 -4y^2 -5 = 0 Collecting like terms: 4y -3y^2 -1 = 0 Divide all terms by -3: y^2 -4/3y +1/3 = 0 Completing the square: (y -2/3)^2 -4/9 +1/3 = 0 => (y -2/3)^2 = 1/9 Square root both sides: y -2/3 = -1/3 or +1/3 Add 2/3 to both sides: y = 1/3 or 1 By substitution into original equation intersection are at: (7/3, 1/3) and (3, 1)
