Your function is not defined at x=0 and x=4 (you have vertical asymptotes there). So, the denominator must be zero at x=0 and x=4. This is the case for
So, the function with that denominator is graphed.
7y+2(5y-9)=16
7y+10y-18=16
17y=34
y=2
5(2)-9=x
10-9=x
x=1
Answer:
Marcus score is 13.7, the mean is 12.89, and the standard deviation is 1.95, so I’ll do ( 13.7 – 12.89)/ 1.95, Since Marcus' z-score is 0.415. The area to the left of the z-score is 65.9% so therefore Marcus did scored better than 65.9% of the students. So Marcus needs to score 98% better of all the other students in order for him to receive an actual certificate.
Step-by-step explanation:
Step-by-step explanation:
<h2>
<em>y = ± 5⁄2 x</em></h2><h2>
<em>y = ± 5⁄2 xThe equation is the left half of the hyperbola. The domain is ( – ∞, – 1⁄5 ]. The range is ( – ∞, ∞ ). The vertical line test indicates that this is not the graph of a function.</em></h2>
