I assume you're asked to solve
4 cos²(<em>x</em>) - 7 cos(<em>x</em>) + 3 = 0
Factor the left side:
(4 cos(<em>x</em>) - 3) (cos(<em>x</em>) - 1) = 0
Then either
4 cos(<em>x</em>) - 3 = 0 <u>or</u> cos(<em>x</em>) - 1 = 0
cos(<em>x</em>) = 3/4 <u>or</u> cos(<em>x</em>) = 1
From the first case, we get
<em>x</em> = cos⁻¹(3/4) + 2<em>nπ</em> <u>or</u> <em>x</em> = -cos⁻¹(3/4) + 2<em>nπ</em>
and from the second,
<em>x</em> = <em>nπ</em>
where <em>n</em> is any integer.
f = amount of fliers expected to be mailed in 1 day.
a)
so, she has 5 days to mail those fliers, so
day 1........... 1f mailed out
day 2........... 2f mailed out
day 3........... 3f mailed out
day 4........... 4f mailed out
day 5........... 5f mailed out
so she'll be mailing out a total of 5f fliers, we know whatever "f" is, that total must be at least 850, now, it could be more, no less, could also be 850 exactly, but could also be more, 5f ⩾ 850.
b)
Answer: slope of the line is 2/3
explanation:
rise over run
go up twice and 3 times to the side
A. NOT
(-5, 2) → x = -5, y = 2
y ≤ x - 5 → 2 ≤ -5 - 5 → 2 ≤ -10 FALSE
B. YES
(5, -2) → x = 5, y = -2
y ≤ x - 5 → -2 ≤ 5 - 5 → -2 ≤ 0 TRUE
y ≥ -x - 4 → -2 ≥ -5 - 4 → -2 ≥ -9 TRUE
C. NOT
(-5, -2) → x = -5, y = -2
y ≤ x - 5 → -2 ≤ -5 - 5 → -2 ≤ -10 FALSE
D. NOT
(5, 2) → x = 5, y = 2
y ≤ x - 5 → 2 ≤ 5 - 5 → 2 ≤ 0 FALSE
