Answer:
Step-by-step explanation:
Hello,
as
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>
A = pi r^2
a = 3.14 * 97^2
a = 29,544.26 mm^2
When two lines intersect, opposite angles are equal. This means that angle 1 equals angle 4. We can use that information to find their values.
Angle 1 = Angle 4
6n+1 = 4n+19
2n=18
n=9
6(9)+1=54+1=55
Angle 1 and 4 equal 55 degrees.
Two angles that form a straight line together have a total sum of 180 degrees. Angles 1 and 5 are like this, as well as Angles 4 and 5, and Angles 4, 3, and 2 added together.
Therefore, 180 = (angle 4) + (angle 3) + (angle 2)
180= 55+(angle 3) + (angle 2)
125= angle 3 + angle 2
I'm not sure what else can be extrapolated from this. There doesn't seem to be a way to find out what the measure of angle 2 is without angle 3 as well. I hope this helps and you can figure it out from the answer choices!
Step-by-step explanation:
x = the total money she had first.
x - x×1/4 - (x - x×1/4)×1/2 = 15
you see, first we needed to deduct 1/4 of x from x. that left us with (x - x×1/4). and from that we need to deduct 1/2.
now, we can simplify :
(x - x×1/4) = (x×4/4 - x×1/4) = x×3/4
x×3/4 - x×3/4 × 1/2 = 15
x×3/4 - x×3/8 = 15
we multiply both sides by 8 (4 is a factor of 8, so this handled both fractions perfectly) :
x×3×2 - x×3 = 120
6x - 3x = 120
3x = 120
x = $40
she had $40 at the beginning.
