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What else can I help you with?
What is y divided by 2 plus y?
If you mean y/2 + y (without parentheses), y/2 is the same as (1/2)y, and y is the same as 1y. You can use the distributive property to combine this; in this case, the result is (3/2)y.
What is the solution of (y-2) (y2)?
If you mean: (y-2)(y+2) then it is y^2 -4 which is the difference of two squares
What makes the equation correct 13 equals y-2?
y = 15 so that 13= 15 - 2 y-2 = 13 y-2 +2 = 13 +2 y = 15
What is the y intercept of y equals 4x - 2?
2
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 the sum of 2x divided by y-2 and 2x divided by 2-y?
2x / (y-2) + 2x / (2-y)First get a common denominator: (y-2) (2-y)Then you have:2x(2-y) / (y-2)(2-y) + 2x(y-2) / (y-2)(2-y), summing and simplifying, you get:(4x-4) / (y-2)(2-y)
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 slope and y-intercept in the equation -2x-y-5?
slope is 2 y intercept is 5/2
What points lie on 3x-y 2?
1
How do you factor x3 - y3?
0
Simplify square root of 36x4y2?
If x is a multiplication symbol, then square root of 36(4)y^2 = square root of(4^2)(2^2)(y^2) = (4)(2)(y) = 8y If it is a variable x, then square root of 36x4y^2 = square root of (4^2)(x)(2^2)(y^2) = (4)(2)(y)(square root of x)= 8y(square root of x)
Quadratic form Factor the polynomial X4 3X2 2?
If you mean x^4 + 3x^2 +2, then you would use any letter. Say y = x^2 therefore x^4 + 3x^2 +2 must = y^2 + 3y + 2 you can now factorise this: (y+2)(y+1) so y = -2 or y = -1 Remember, y = x^2 so, -2 = x^2. square root this. But if you remember what you've been taught, then you cannot square root a (-). So, there are no solutions. This means that the line on the graph will not intersect the x-axis. I hope this is what you are looking for.
Answers to a math test?
3(4-y) = 5(y+2), what does y =
Graph y equals -2?
A straight horizontal line passing through y=-2
How do you solve for x when y equals quantity x plus 2 divided by quantity x plus 10?
y=(x+2)/(x+10) xy+10y=x+2 xy-x=2-10y x(y-1)=2-10y x=(2-10y)/(y-1)
How do you find a perpendicular line?
Given y = 2x the slope of that line is simply 2, A perpendicular line would have the negative inverse of 2, -1/2 or -.5 thus y = 2x is perpendicular to y = -.5x Y = -(1/Mx) + b is perpendicular to: Y= Mx + b
Who are some geniuses?
PIERRE DE FERMAT' S LAST THEOREM. CASE SPECIAL N=3 AND.GENERAL CASE N>2. . THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2. Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N. SPECIAL CASE N=3. WE HAVE (X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2. BECAUSE X*Y>0=>2X^2*Y^2>0. SO (X^2+Y^2)^2=/=X^4+Y^4. CASE 1. IF Z^2=X^2+Y^2 SO (Z^2)^2=(X^2+Y^2)^2 BECAUSE (X^+Y^2)^2=/=X^4+Y^4. SO (Z^2)^2=/=X^4+Y^4. SO Z^4=/=X^4+Y^4. CASE 2. IF Z^4=X^4+Y^4 BECAUSE X^4+Y^4.=/= (X^2+Y^2.)^2 SO Z^4=/=(X^2+Y^2.)^2 SO (Z^2)^2=/=(X^2+Y^2.)^2 SO Z^2=/=X^2+Y^2. (1) AND (2)=> Z^4+Z^2=/=X^4+Y^4+X^2+Y^2. SO 2Z^4+2Z^2=/=2X^4+2Y^4+2X^2+Y^2. SO (Z^4+Z^2+2Z^3+Z^4+Z^2-2Z^3)=/=(X^4+X^2+2X^3+X^4+X^2-2X^3)+)(Y^4+Y^2+2Y^3+Y^4+Y^2-2Y^3) SO IF (Z^4+Z^2+2Z^3)/4=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4 => (Z^4+Z^2-2Z^3)/4=/=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3/4) AND SO IF (Z^4+Z^2-2Z^3)/4=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3)./4 => (Z^4+Z^2+2Z^3)/4=/=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4 BECAUSE (Z^4+Z^2+2Z^3)/4 - (Z^4+Z^2-2Z^3)/4 =Z^3. SO Z^3=/=X^3+Y^3. GENERAL CASE N>2. Z^N=/=X^N+Y^N. WE HAVE [X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=X^(N+1)/2+Y^(N+1)/2+ H. BECAUSE X*Y>0=>H>0. SO [X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=/= X^(N+1)/2+Y^(N+1)/2 CASE 1. IF Z^(N-1)/2=X^(N-1)/2+Y^(N-1)/2 SO [Z^(N-1)/2]^(N+1)/(N-1)=[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1). BECAUSE [X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2. SO [Z^(N-1)/2]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2. SO Z^(N+1)/2=/=X^(N+1)/2+Y^(N+1)/2. CASE 2. IF Z^(N+1)/2=X^(N+1)/2+Y^(N+1)/2 SO [Z^(N+1)/2]^(N-1)/(N+1)=[X^(N+1)/2+Y^(N+1)/2 ]^(N-1)/(N+1) BECAUSE [X^(N+1)/2+Y^(N+1)/2](N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2. SO [Z^(N+1)/2]^(N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2. SO Z^(N-1)/2=/=X(N-1)/2+Y^(N-1)/2.. SO (1) AND (2)=> Z^(N+1)/2+Z^(N-1)/2=/=X^(N+1)/2+Y^(N+1)/2+X^(N-1)/2+Y^(N-1)/2. SO 2[Z^(N+1)/2+Z^(N-1)/2]=/=2[X^(N+1)/2+Y^(N+1)/2]+2[X^(N-1)/2+Y^(N-1)/2.] SO [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]+[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]+[X^(N+1)/2+X^(N-1)/2-2X^N ]+[Y^(N+1)/2+Y^(N-1)/2+2Y^N ]+[Y^(N+1)/2+Y^(N-1)/2-2Y^N ] SO IF [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2+2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4=> [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4 AND IF [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4 => [Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]/4 + [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4 BECAUSE [Z^(N+1)/2+Z^(N-1)/2+2Z^N ] /4- [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=Z^N. SO Z^N=/=X^N+Y^N HAPPY&PEACE. Trantancuong.
