Answer:
your missed answer is V 62.1
Step-by-step explanation:
Complete Question
The complete Question is attached below
Answer:
Option D
Step-by-step explanation:
From the question we are told that:
Slant asymptote of 
Vertical asymptote at 
Points (0,6)
Generally the Denominator is give as
With
Vertical Asymptote at
Therefore
Denominator = (x-1)
Generally Slant asympote 2x Gives the Coefficient of the numerator
Therefore
The expression for a rational function f(x) that satisfies the conditions
Option D
Answer:
Note: The expression "h =300- pi r /pi r" would equate to 299, as written. I find h = (300 - π
)/πr to be the equation for height.
Step-by-step explanation:
The total surface area of a sylinder is 2πr(h+r) where h is the height and r is the radius. That consists of top and bottom disks, each with an area of π, for a total of 2π. The side is a continous sheet that has an area of 2πrh, where h is the height. 2πr is the circumference of the can, so the surface area of the side is 2πrh.
The total surace area of a closed cylinder is therefore 2π + 2πrh.
2π + 2πrh = 600
π + πrh = 300
πrh = 300 - π
h = (300 - π )/πr
Answer:
Mr. Winking 87.9%
Ms. Sand 88.55%
Ms. Sand's class is better
Step-by-step explanation:
Step 1: Average in Mr. Winking's Class 87.9%
Convert each weight factor to a decimal.
Homework 10% = 0.1
Quiz 25% = 0.25
Test 45% = 0.45
Exam 20% = 0.2
Multiply your grades by the weight factors:
Homework 0.1 X 95% = 9.5%
Quiz 0.25 X 85% = 21.25%
Test 0.45 X 87% = 39.15%
Exam 0.2 X 90% = 18%
Add all of these values:
9.5% + 21.25% + 39.15% + 18%
= 87.9%
Step 2: Average in Ms. Sand’s Class 88.55%
Convert each weight factor to a decimal.
Homework 15% = 0.15
Quiz 20% = 0.2
Test 40% = 0.4
Exam 25% = 0.25
Multiply the weight factors by your grades:
Homework 0.15 X 95% = 14.25
Quiz 0.2 X 85% = 17
Test 0.4 X 87% = 34.8
Exam 0.25 X 90% = 22.5
All all these values:
14.25 + 17 + 34.8 + 22.5
= 88.55%
Step 3:
I would rather be in Ms. Sand's class. My average in her class would be 88.55%, whereas in Mr.Winking's Class, my average would be 87.9%. 88.55% is a higher mark than 87.9% and I want higher marks. Therefore, I would get higher marks in Ms. Sand's class.
