There seems to be a flaw with this question because it says that there are five x-intercepts but the given information only gives you 4 x-intercepts to work with.
Even means the graph is symmetric about the y-axis
The best answer is <span>A.(–6, 0), (–2, 0), and (0, 0)
because you do not have to worry about another point (0,0). Plus we need (-6,0) for it to be symmetric with (6,0).
Consider function f(x) = x²(x-6)(x+6)(x+2)</span>²(x-2)<span>². It is even and fits these conditions as it has x-intercepts at (6,0), (-6,0), (-2,0), (2,0), and (0,0). again, the question does not tell us the fifth x-intercept, so we need to assume that there is another one that needs to be there...and so (-2,0) must have (2,0) for it to be even as well.</span>
The equation given in the question has two unknown variables in the form of "x" and "y". The exact value of "x" and "y" cannot be determined as two equations are needed to get to the exact values of "x" and "y". This equation can definitely be used to show the way for determining the values of "x" in terms of "y"and the value of "y" in terms of "x". Now let us check the equation given.
2x - 5y = - 15
2x = 5y - 15
2x = 5(y - 3)
x = [5(y - 3)]/2
Similarly the way the value of y can be determined in terms of "x" can also be shown.
2x - 5y = - 15
-5y = - 2x - 15
-5y = -(2x + 15)
5y = 2x + 15
y = (2x +15)/5
= (2x/5) + (15/5)
= (2x/5) + 3
So the final value of x is [5(y -3)]/2 and the value of y is (2x/5) + 3.
Answer:
SAS requires two congruent sides and the included angle be also congruent
Given is the picture are congruent triangles
<u>ΔACB ≅ ΔECD, because:</u>
- AC ≅ EC, given
- BC ≅ DC, given
- ∠ACB ≅ ∠ECD, vertical angles
Use the least common denominator (LCD) to write these fractions as equivalent fractions with like denominators, and then compare them two at a time
