0
15y2 + 45y - 9 = 4y - 5y2; thus,
20y2 + 41y - 9 = (5y - 1)(4y + 9) = 0.
Therefore, the solution to the equation given is,
y = 1/5 or -2¼.
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What is 10y squared minus 15y squared?
10y2 - 15y2 = -5y2
What is the recoprocal of 5y2 plus 9y?
1/5y2+9
Simplify the expression 4x2-8y plus 3x-9x2 plus 10y?
-5y2+3x+2y
Factor 5y2-22y plus 8?
That is the most simplifiable version of this equation.
What is the solution when y equals 2x plus 1 is a tangent to the circle 5y2 plus 5x2 equals 1?
If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
What is the greatest common factor of 20y3 5y2 45y?
The GCF is 5y.20y3 = (5y)(4y2)5y2 = (5y)(y)45y = (5y)(9)
What is 10y squared minus 15y squared?
10y2 - 15y2 = -5y2
How do you factor out x squared plus 8xy plus 15y squared?
x2 + 8xy + 15y2 = (x + 3y) (x + 5y)
What is the recoprocal of 5y2 plus 9y?
1/5y2+9
X equals 5y2-20y plus 15?
x isn't in the calculation so it cant be solved.
Simplify the expression 4x2-8y plus 3x-9x2 plus 10y?
-5y2+3x+2y
Factor 5y2-22y plus 8?
That is the most simplifiable version of this equation.
What is the solution when y equals 2x plus 1 is a tangent to the circle 5y2 plus 5x2 equals 1?
If y = 2x+1 is a tangent line to the circle 5y^2 +5x^2 = 1 then the point of contact is at (-2/5, 1/5) because it has equal roots
(-3y2 and minus 8) and minus (-5y2 plus 1)?
it's MINUS 182.5
How do you solve 3.5y2 plus 2.6y-8.2 equals 4y2-6.9y using the quadratic formula?
Multiply each item by 10 to give whole numbers then move items to give ay2+by+c=0 3.5y2+2.6y-8.2=4y2-6.9y multiply by 10 35y2 + 26y - 82 = 40y2 -69y subtract 35y2 from both sides 26y - 82=5y2 -69y subtract 26y from both sides -82 = 5y2 - 95y add 82 to each side 0 = 5y2 - 95y +82 or 5y2 - 95y +82 = 0
Factor the trinomial 9x2 - 18xy plus 5y2 and Factor the trinomial 4x2 plus 24x plus 11?
(3x - y)(3x - 5y) and (2x + 1)(2x + 11)
4x-5y2 12x8 plus 15y solve by elimination?
Without any equality signs the given terms can't be considered as simultaneous equations and so therefore no solutions are possible.
