A great circle is a section of a sphere that passes through its center. If the earth were a sphere, a great circle would be the equator and its axis would be the line connecting the geographic north and south pole. The length of the axis is then equal to the diameter of the sphere. For this problem, the radius of the sphere is 12 inches. A section is formed by slicing through the sphere and all sections of a sphere are circles. Considering the plane to be cut above and parallel with the equator (which is a great circle), the distance of the plane from the center of the sphere would then be the distance between the centers of the sphere and section. It is also given that the radius of the section is 9 inches. A right triangle is formed by connecting the center of the sphere, an edge of the section, and back to the center of the sphere whose hypotenuse is 12 inches (radius of the sphere), one leg is the 9 inches (radius of the section), and another leg is the distance of the plane from the sphere's center. Thus, the distance can be calculated using the Pythagorean theorem, d = sqrt(12^2 - 9^2) = sqrt(144 - 81) = sqrt(63) = 3*sqrt(7).
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Answers:
121/11=11
23/3=23/3
85/5=17
Well, you have to be given an interval. That is not an interval.
You should be given something like
If you add those fractions up you'll just get a point on the number line.
Find the LCD for all of them which is 9/6 - 4/6 + 3/6 = 8/6 = 2/3
You'll end up with a point on 2/3
Answer:
f(2n)-f(n)=log2
b.lg(lg2+lgn)-lglgn
c. f(2n)/f(n)=2
d.2nlg2+nlgn
e.f(2n)/(n)=4
f.f(2n)/f(n)=8
g. f(2n)/f(n)=2
Step-by-step explanation:
What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]
from a
f(n)=logn
f(2n)=lg(2n)
f(2n)-f(n)=log2n-logn
lo(2*n)=lg2+lgn-lgn
f(2n)-f(n)=lg2+lgn-lgn
f(2n)-f(n)=log2
2.f(n)=lglgn
F(2n)=lglg2n
f(2n)-f(n)=lglg2n-lglgn
lg2n=lg2+lgn
lg(lg2+lgn)-lglgn
3.f(n)=100n
f(2n)=100(2n)
f(2n)/f(n)=200n/100n
f(2n)/f(n)=2
the time will double
4.f(n)=nlgn
f(2n)=2nlg2n
f(2n)-f(n)=2nlg2n-nlgn
f(2n)-f(n)=2n(lg2+lgn)-nlgn
2nLg2+2nlgn-nlgn
2nlg2+nlgn
5.we shall look for the ratio
f(n)=n^2
f(2n)=2n^2
f(2n)/(n)=2n^2/n^2
f(2n)/(n)=4n^2/n^2
f(2n)/(n)=4
the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare
6.n^3
f(n)=n^3
f(2n)=(2n)^3
f(2n)/f(n)=(2n)^3/n^3
f(2n)/f(n)=8
the ratio will be times 8 the initial
7.2n
f(n)=2n
f(2n)=2(2n)
f(2n)/f(n)=2(2n)/2n
f(2n)/f(n)=2
Answer:
undefined
The problem:
Find the slope for the line going through (3,4) and (3,-4).
Step-by-step explanation:
Line up points vertically and subtract.
Then put 2nd difference over 1st.
( 3 , 4)
-(3 , -4)
------------
0 8
So the slope would have been 8/0 but this is undefined.
So the slope is undefined.
Also notice the x's are the same and the y's are difference so this is a vertical line. There is only rise in a vertical line and no run. Recall, slope is rise/run. You cannot divide by 0 so this is why we say the slope is undefined when the x's are always the same no matter the y.
