Factor the denominator:
(3x-1)(x+2)
every time x=-2 (because of (x+2), it would cause a warp in the graph, and not have an input. that is the vertical asymptote.
since you have factor 3x-1 on top and bottom, in which the zero is 1/3, there would be no solution at x=1/3. it would be a hole.
the answer here is C.
Answer: ![\begin{bmatrix}\mathrm{Solution:}\:&\:x\le \frac{17}{3}\:\\ \:\mathrm{Decimal:}&\:x\le \:5.66666\dots \\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:\frac{17}{3}]\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Ax%5Cle%20%5Cfrac%7B17%7D%7B3%7D%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BDecimal%3A%7D%26%5C%3Ax%5Cle%20%5C%3A5.66666%5Cdots%20%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%28-%5Cinfty%20%5C%3A%2C%5C%3A%5Cfrac%7B17%7D%7B3%7D%5D%5Cend%7Bbmatrix%7D)
Step-by-step explanation:






The answer is 8.33333333 I believe
Answer:
$59.4
Step-by-step explanation:
Round up: 60*1 = $60. Everytime you add $0.99, you subtract off a penny, so you will subtract off 60 pennies. 60 cents from the original answer, so 60-0.6=59.4.
Answer:
1/8
Step-by-step explanation:
sin²(π/8) − cos⁴(3π/8)
Use power reduction formulas:
1/2 (1 − cos(2×π/8)) − 1/8 (3 + 4 cos(2×3π/8) + cos(4×3π/8))
Simplify:
1/2 (1 − cos(π/4)) − 1/8 (3 + 4 cos(3π/4) + cos(3π/2))
1/2 (1 − √2/2) − 1/8 (3 + 4 (-√2/2) + 0)
1/2 − √2/4 − 1/8 (3 − 2√2)
1/2 − √2/4 − 3/8 +√2/4
1/2 − 3/8
1/8