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What is a cubic trinomial?

A cubic trinomial is a polynomial expression that consists of three terms and has a degree of three. It typically takes the form ( ax^3 + bx^2 + cx ), where ( a ), ( b ), and ( c ) are coefficients, and ( a ) is non-zero. The expression can represent various algebraic relationships and is often used in polynomial equations and functions.


Is this a trinomial square x2-5x 25?

The expression (x^2 - 5x + 25) is not a trinomial square. A trinomial square takes the form ((a - b)^2 = a^2 - 2ab + b^2), which would include a linear term with a coefficient that is double the product of (a) and (b). In this case, the constant term (25) is not the square of half the coefficient of (x) (which would be ((-\frac{5}{2})^2 = \frac{25}{4})). Thus, this expression does not fit the criteria for a trinomial square.


How do you describe a perfect square trinomial?

A perfect square trinomial is a quadratic expression that can be expressed as the square of a binomial. It takes the form (a^2 + 2ab + b^2) or (a^2 - 2ab + b^2), where (a) and (b) are real numbers. The resulting trinomial can be factored as ((a + b)^2) or ((a - b)^2). This characteristic makes perfect square trinomials particularly useful in algebra for solving equations and simplifying expressions.


Is a perfect square trinomial of the form a2-2ab b2 16x2-36x 9 true or false?

False. A perfect square trinomial takes the form ( (a - b)^2 = a^2 - 2ab + b^2 ) or ( (a + b)^2 = a^2 + 2ab + b^2 ). The expression ( 16x^2 - 36x + 9 ) can be factored as ( (4x - 3)^2 ), which confirms it is a perfect square trinomial, but it does not fit the specified form ( a^2 - 2ab b^2 ).


When a trinomial is factored as (x a)(x b) what is the sum of a and b?

Thanks to the rubbish browser which you are required to use for posting questions, we cannot see most mathematical symbols. It is, therefore, not possible to give an unambiguous answer.The algebraic sum of a and b is the coefficient of the middle term - the term for x.Incidentally, the algebraic sum is the sum of a and b which takes account of their signs.


Is it right to takes sides when your are both arguing and your stuck in the middle?

No


What does the flock first live in Maximum Ride?

The School is hidden in Death Valley, so the first half of the book takes place in southwest America. In the middle of the book, the Flock fly to New York City.


What is the motto of Rudder Middle School?

The motto of Rudder Middle School is 'Rudder kids are worth whatever it takes.'.


Why liquids have two coefficients of expensions?

liquid has two coefficient of expensionbecause during expension by heating the expension of container as well as expension of liquid takes place. that is why liquid have two coefficient of expension


How do you get to bubble lake on Mario and luigi inside bowser story?

Go through the forest and on the left side in the middle of the forest is a tube that takes you to the first stage in bubble lake.


What is a ruler who takes power with the support of the middle and working classes?

Tyrant


How do you factor 12x2 plus 28x-17?

The trick with this kind of question is to find two terms that add up to the middle one (28x), and whose coefficients are equal multiples of the first and last term: 12x2 + 28x - 17 = 12x2 - 6x + 34x - 17 = 6x(2x - 1) + 17(2x - 1) = (6x + 17)(2x - 1) The only trick with this technique is that sometimes it's difficult to find those two middle terms (-6x and 34x in this case). If it takes too long to figure this out, you can actually break it down into ~another~ quadratic equation, which will usually be easier to factor, and give you those two values. Consider: Let "a" and "b" be the unknown coefficients. We know: b = 28 - a <-- we know this because they must add up to the middle term 12/a = b/17 <-- this must be true in order for us to end up with a common multiple. ∴ 12/a = (28 - a)/17 ∴ 28a - a2 = 204 ∴ a2 - 28a + 204 = 0 We can then solve for "a" by factoring this equation: (a - 34)(a + 6) = 0 So our coefficients are the two possible values for "a", 34 and -6. We can then simply plug them in to the original equation and factor it out as shown above.