Think of numbers that when they are divided by 10, the remainder is 9.
So numbers like: 19, 29, 39, 49, 59, 69, 79, 89, and 99.
Now numbers that when they are divided by 9, the remainder is 8.
So numbers like: 17, 26, 35, 44, 53, 62, 71, 80, 89, and 98.
Now the number that is alike from the two sets of numbers is n.
n=89
Double check your work.
Divide 89 by 10. Remainder of 9
Divide 89 by 9. Remainder of 8.
Hope this helps :)
For me personally, the easiest way to do this is by isolating the x² term, and finding the square root of both sides. The hardest way (well actually, the longest way) would be to use the quadratic formula. It just complicates things unnecessarily.
Answer:
independent: day number; dependent: hours of daylight
d(t) = 12.133 +2.883sin(2π(t-80)/365.25)
1.79 fewer hours on Feb 10
Step-by-step explanation:
a) The independent variable is the day number of the year (t), and the dependent variable is daylight hours (d).
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b) The average value of the sinusoidal function for daylight hours is given as 12 hours, 8 minutes, about 12.133 hours. The amplitude of the function is given as 2 hours 53 minutes, about 2.883 hours. Without too much error, we can assume the year length is 365.25 days, so that is the period of the function,
March 21 is day 80 of the year, so that will be the horizontal offset of the function. Putting these values into the form ...
d(t) = (average value) +(amplitude)sin(2π/(period)·(t -offset days))
d(t) = 12.133 +2.883sin(2π(t-80)/365.25)
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c) d(41) = 10.34, so February 10 will have ...
12.13 -10.34 = 1.79
hours less daylight.
Answer:
Step-by-step explanation:
To make the problem easier to solve, we will set it up as the equation of the length of time of each class times the number of classes equals the total amount of minutes. However, since we don't know the number of classes, we'll symbolize our two unknowns with two variables.
75x + 45y = 705
(75x + 45y)/15 = 705/15
5x + 3y = 47
y = (47-5x)/3
It looks like we can't simplify the equation any more, so now it is a matter of trial and error. The minimum number of Saturday classes means the maximum number of weekday classes. We first will test for the maximum by assuming there are no Saturday classes, then will work our way up until x is an integer.
If x = 0
(47-5(0))/3 = 47/3 = 15.6666
If x = 1
(47-5(1))/3 = 42/3 = 14
This works. Therefore, the maximum number of weekday classes is 14, or choice b.
Answer: D
<u>Step-by-step explanation:</u>
Vertex form is: f(x) = a(x - h)² + k where (h, k) is the vertex.
f(x) = (x - 3)² - 5
- the vertex is (3, 5).
- "a" is positive (+1), so it points upward.
The only graph that matches is the last one, <em>which I called graph D.</em>