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6-4y = 2
What else can I help you with?
Divide 40x over 64y by 5x over 8y?
140x / 64y divide 5x/8y = > 40x/64y X 8y/5x Note the change of sign and the inversion of the Right hand fraction. Cancel down by '5x' 8/64y X 8y/1 Cancel down by '8y' 8/8 X 1/1 = 8/8 = 1 The answer!!!!
What is the product of (9x-8y)(9x 8y)?
If you mean (9x-8y)(9x+8y) then it is 81x^2 minus 64y^2
Is y6 plus 8y3 plus 64y a trinomial with a constant term?
No.
What is 64y cubed plus 1?
You can't simplify it any further without a value for y.
How do you solve 81x2 144xy 64y2?
To solve the expression ( 81x^2 + 144xy + 64y^2 ), you can recognize it as a perfect square trinomial. It can be factored as ( (9x + 8y)^2 ). Thus, the solution is ( (9x + 8y)^2 ).
X plus 8y equals 22 -8x-2y plus 10?
8x + 64y = 176 Add: 62y = 186 so y = 3 and x = -2 This assumes you mean "equals 10", not "plus 10"
How do I solve for y the linear equations 8x-7y equals 7 and 7x equals 8y equals 8?
8x - 7y = 7 (A)7x + 8y = 8 (B)7*(A): 56x - 49y = 498*(B): 56x + 64y = 648*(B)-7*(A): (64+49)*y = 64-49=> 113y = 15=> y = 15/113
What two numbers are the product of 900 but there sum equal 64?
ANSWER: The two numbers required are not whole numbers. These are 43.135 and 20.865.Solution 1:Assuming you want whole numbers:Prime factorization for 900 is2 x 2 x 3 x 3 x 5 x 5You cannot combine these in any way to have a sum that is 64. You can with 65, but not 64.Solution 2:Let:x = first numbery = second numberx * y = 900 (Equation 1)x + y = 64x = 64 - y (Equation 2)Substitute Eq. (2) in Eq. (1):x * y = 900(64 - y) y = 90064y - y2 = 900- y2 +64y - 900 = 0 (Equation 3)multiply Eq. (3) by -1y2 - 64y + 900 = 0using quadratic equation:y1 = 43.135 y2 = 20.865Therefore, the two numbers are 43.135 and 20.865
What two numbers multipy to equal 900 and the sum of the two numbers equals 64?
Solution 1: Prime factorization for 900 is2 x 2 x 3 x 3 x 5 x 5You cannot combine these in any way to have a sum that is 64. You can with 65, but not 64.Solution 2:Let:x = first numbery = second numberx * y = 900 (Equation 1)x + y = 64x = 64 - y (Equation 2)Substitute Eq. (2) in Eq. (1):x * y = 900(64 - y) y = 90064y - y2 = 900- y2 +64y - 900 = 0 (Equation 3)multiply Eq. (3) by -1y2 - 64y + 900 = 0using quadratic equation:y1 = 43.135 y2 = 20.865Therefore, the two numbers are 43.135 and 20.865
Find a possible formula for the quadratic function whose graph has vertex left parenthesis 8 comma 2 right parenthesis and y intercept negative 11?
Using the vertex (h, k) the equation of the quadratic would be: y = a(x - h)² + k → y = a(x - 8)² + 2 → y = a(x² - 16x + 64) + 2 → y = ax² -16ax + 64a + 2 Using the y-intercept of (0, -11): -11 = a 0² - 16a 0 + 64a + 2 → 64a = -13 → a = -13/64 Thus the quadratic could be: y = (-13/64)x² - 16(-13/64)x + 64(-13/64) + 2 → y = (-13/64)x² + (13/4)x - 11 Or, removing the fractions: 64y = -13x² + 208x - 704
What is the complete factor of x6-64y6?
The expression ( x^6 - 64y^6 ) can be recognized as a difference of squares, which can be factored as ( (x^3 - 8y^3)(x^3 + 8y^3) ). Each of these terms can be further factored using the sum and difference of cubes formulas. Thus, the complete factorization is ( (x - 2y)(x^2 + 2xy + 4y^2)(x + 2y)(x^2 - 2xy + 4y^2) ).
How do you foil -y32x2y-4xy2?
You can't exactly FOIL it because you don't have a problem in the form of (3x + 2)(4x - 6) Frankly, I don't understand exactly what question you are trying to solve, so I'll give you a couple of examples that might help. (-y32) times (2y - 4xy2) turns into: ((-1)(y32) times 2y) plus ((-1)(y32) times ((-1)(4xy2)) which becomes -2y33 plus 4xy34 which would be more often expressed as 4xy34 - 2y33 Perhaps you meant something more like this: (-y times 32) times (2y - 4xy2) turns into: (-32y) times (2y - 4xy2) which becomes (-y times 2y) + (-y times -4xy2) + (32 times 2y) + (32 times -4xy2) which becomes -2y2 + 4xy3 + 64y - 128xy2 which is 4xy3 -128 xy2 - 2y2 + 64y I'm sure if I'm gotten it wrong, someone will straighten it out. Also, if this doesn't help, please fix up the original question so it can be answered properly.
