Multiply every term in both equations by any number that is not 0 or 1, and has not been posted in our discussion already. Then solve the new system you have created using elimination or substitution method:6x + 9y = -310x - 6y = 58
y=x2-18x+52First off, this does not factor cleanly, but even if it did it would not help us. We must complete the square to yield a perfect-square trinomial (a trinomial that can be written in the form (x-a)2 where a is a real number.To do this, halve the coefficient of the single-x term (for this case, the term 18x) and square it.18/2=992=81The result of this process can be added to the two terms involving x to get a perfect square trinomial. For instance, the trinomial that would result would be:x2-18x+81=(x-9)2However, there is already 52 of that 81 that is needed present, so in reality, you only need to add 29 (81-52=29) to the right side of the equation to get a perfect-square trinomial. Whatever is done to one side of the equation must be done to the other side, so since you added 29 to the right side, you must add 29 to the left side, giving you:y+29=x2-18x+52+29y+29=x2-18x+81y+29=(x-9)2y=(x-9)2-29This is in vertex form, you can see that the vertex is at (9,-29)
The terms consistent and dependent are two ways to describe a system of linear equations. A system of linear equations is dependent if you can algebraically derive one of the equations from one or more of the other equations. A system of linear equations is consistent if they have a common solution.An example of a dependent system of linear equations:2x + 4y = 84x + 8y = 16Solve the first equation for x:x = 4 - 2yPlug that value of x into the second equation:16 - 8y + 8y = 16, which gives 16 = 16.No new information was gained from the second equation, because we already knew 16 = 16, so these two equations are dependent.An example of an inconsistent system of linear equations:Because consistency is boring.2x + 4y = 84x + 8y = 15Solve the first equation for x:x = 4 - 2yPlug that value of x into the second equation:16 - 8y + 8y = 15, which gives 16 = 15.This is a contradiction, because 16 doesn't equal 15. Therefore this system has no solution and is inconsistent.
Typically, if a list of items has semicolons separating each item, commas are not required.
Just to get you straight, that is actually an expression. Equations have equal signs in them. Just so you know. Anyway, it just so happens that this expression (-6n - 30) is already simplified. There are no like terms to combine there, since 30 does not have an "n".
To insert a quadratic formula (or any other scientific formula) into a Word document, go toInsert (tab) > Equations (under the Symbols block)From there you can either select the format of the formula you would like to insert if a template is available (there is a template already for quadratic equations) but if there isn't one, can either download on from Office.com OR create your own by clicking Insert New Equation.
Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.
The phrase "other ways" implies you already have one way in mind. I regret that I am unable to read your mind to determine what that way might be!
Well, that's one method to solve the quadratic equation. Here is an example (using the symbol "^" for power): solve x^2 - 5x + 6 = 0 Step 1: Convert the equation to a form in which the right side is equal to zero. (Already done in this example.) Step 2: Factor the left side. In this case, (x - 3) (x - 2) = 0 Step 3: Use the fact that if a product is zero, at least one of its factors must be zero. This lets you convert the equation to two equations; x - 3 = 0 OR x - 2 = 0 Step 4: Solve each of the two equations.
If you already know that x = -3 and y = 5 what linear equations are you wanting to solve?
Your question already contains the answer. A satirist writes satires.
coffee or anything that contains caffene
Multiply every term in both equations by any number that is not 0 or 1, and has not been posted in our discussion already. Then solve the new system you have created using elimination or substitution method:6x + 9y = -310x - 6y = 58
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Since 0.83 contains two decimal places, it is already rounded to the nearest hundredth.