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First, the standard form of a quadratic equation is
ax2 + bx + c = 0 a ≠ 0
Here the only needs are that we have an x2 in the equation. We guarantee that this term will be exist in the equation by needing a ≠ 0 . However, Note that it is okay if b and/or c are zero.
There are several ways to solve quadratic equations. The first two methods won't always work, yet are possibly a little simpler to use while they work. In This section we will cover with these two methods.
25/30=x/12
Now we will discuss as solving logarithmic equations, or equations along with logarithms in them. We will be looking at two particular types of equations here. In specific we will
I do not know how to do this
A student rented a bicycle for a one-time fee of $12.00 and then a charge of $0.85 per day.She paid $28.15 for the use of the bicycle. How many days did she keep it?
How can x raised to the second power mines x mines 2 . can be factored as (x+1)(x-2)
are these like terms
y = 4 - 3x /1 + 8x for x. Solution This one is very alike to the previous instance. Here is the work for this problem. y + 8xy = 4 - 3x 8xy + 3x = 4 - y X(8 y +3)
-6k+7k
3y+2.5x+3.4 use graphing calculator and the x intercept approach. Use window[-3,3,1][-3,3,1]
which of the following are cyclic group G1= G2= G3= G4= G5={6n/n belong to z}
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