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There is no equation (nor inequality) in the question, not even a valid expression. So there is nothing that can be solved.
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∙ 10y ago
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What is a is a trinomial expression?
aX^2 + bX + c
What is an Quadratic in form?
Ax^2 + bx + c = 0
What is standard the form of a quadratic equation?
Ax 2+Bx+c=0
How do you find the x-intercepts of this function parabola?
Solve the quadratic equation.If y = ax^2+bx+c then the intercepts arex = [-b +/- sqrt(b^2 - 4ac)]/(2a).The solutions are real if and only if b^2 >= 4ac
How do you complete the square for negative quadratics I.e. in the form -ax2 bx c?
Without using complex numbers, which is a bad idea, you can't. But you can do something which is almost as good: instead of dealing with -ax^2+bx+c, you can take -(ax^2-bx-c), and then complete the square for the inside quadratic. This will give you something like -((rx+s)^2+t), which is usually good enough. For example, if you're completing the square in order to solve a quadratic equation, the sign on the outside doesn't matter: if -((rx+s)^2+t)=0 then (rx+s)^2+t=0 as well (because the negative of 0 is 0). A different approach, which is usually more useful, is to factor out the a, regardless of whether it's positive or negative. Then instead of completing the square for ax^2+bx+c, you complete the square for x^2+b/ax+c/a, and multiply the final result by a.
What is the answer of factoring by grouping terms of ax plus 2a plus bx plus 2b?
(x + 2)(a + b)
What is the integral of sin3ycos5ydy?
Best way: Use angle addition. Sin(Ax)Cos(Bx) = (1/2) [sin[sum x] + sin[dif x]], where sum = A+B and dif = A-B To show this, Sin(Ax)Cos(Bx) = (1/2) [sin[(A+B) x] + sin[(A-B) x]] = (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] Using the facts that cos[-k] = cos[k] and sin[-k] = -sin[k], we have: (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[Bx]-sin[Bx]cos[Ax])] (1/2) 2sin[Ax]Cos[Bx] sin[Ax]Cos[Bx] So, Int[Sin(3y)Cos(5y)dy] = (1/2)Int[Sin(8y)-Sin(2y)dy] = (-1/16) Cos[8y] +1/4 Cos[2y] + C You would get the same result if you used integration by parts twice and played around with trig identities.
How do you solve for roots in a parabola?
If the equation of the parabola isy = ax^2 + bx + c then the roots are [-b +/- sqrt(b^2-4ac)]/(2a)
What is a is a trinomial expression?
aX^2 + bX + c
What is an Quadratic in form?
Ax^2 + bx + c = 0
Ax2 bx c0?
ax^2+bx+c=0 is the standard form of a quadratic function.
How do you solve the equation ax bxc?
The expression you presented is not an equation. Do you mean ax2 + bx = c? Do you mean to solve it for x? I'm assuming that's the case, but you need to be more clear on your question. To solve for x then, the technique to use is called completing the square: ax2 + bx = c Multiply both sides by a: a2x2 + abx = ac Add the square of b/2 to both sides: a2x2 + abx + (b/2)2 = ac + (b/2)2 We now have a perfect square on the left, simplify: (ax + b/2)2 = ac + b2 / 4 (ax + b/2)2 = (4ac + b2) / 4 And now solve for x: ax + b/2 = ±[(4ac + b2) / 4]1/2 ax + b/2 = ± √(4ac + b2) / 2 ax = [-b ± √(4ac + b2)] / 2 x = [-b ± √(4ac + b2)] / 2a
Find the x value for point C such that AC and BC form a 23 ratio.?
It is (2*Ax + 3*Bx)/5 where Ax and Bx are the x coordinates of A and B.
8086 program for addition of two 64 bit numbers?
8150
What is standard the form of a quadratic equation?
Ax 2+Bx+c=0
How do you write a parabola in standard form?
y = ax^2 + bx + c
What is the equation for a parabola?
The general equation for a parabola is y = ax^2 + bx + c, where a, b, and c are constants that determine the shape, orientation, and position of the parabola.
