find the circumference of the circle then multiply it by 24
Answer:
Become the leader of our nation.
Step-by-step explanation:
Like all Nazi propaganda, the 1936 poster suggests that young Germans are highly capable of becoming leaders of the nation. A young man who must be blond, with teeth and a perfect body. It was unconditional for Nazi propagandists to convince young people to support the goals and policies of the political party.
On the other hand, the leaders of the Nazi youths sought to integrate the children into the national Nazi community and prepare them to serve as soldiers in the armed forces or, later, in the SS.
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.
So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))
now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16
Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84
So we have our linear approximation for the two.
If you wanted to, you could check your answer by finding g(x). Since you know g'(x), take the antiderivative and we will get
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35
So just to check our linear approximations use that to find g(2.99) and g(3.01)
g(2.99) = -5.1597
g(3.01) = -4.8397
So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer. Not a bad method if you ever need to use it.
Answer:
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9 3(0) ≥ (0) - 9
3 ≥ - 9
yes 3x + y > - 3 3(0) + (0) > - 3
3 > - 3
yes so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5
yes3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3
yesSo the graph is the bottom right one since (4, 2) is included in that shaded region.