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Juli2301 [7.4K]
3 years ago
14

What are some examples of implicit memory?

Advanced Placement (AP)
2 answers:
horrorfan [7]3 years ago
5 0
Examples of implicit memory include: Remembering the words to a song and finishing a line of a song when someone sings the first two words. Recalling how to walk, ride a bike, and engage in other procedural tasks.
Tomtit [17]3 years ago
3 0
It is a type of long term memory that stands in contrast to explicit memory in that it doesnt require conscious thought.
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The Berlin Conference of 1884–1885 marked the climax of the European competition for territory in Africa, a process commonly known as the Scramble for Africa. During the 1870's and early 1880's European nations such as Great Britain, France, and Germany began looking to Africa for natural resources for their growing industrial sectors as well as a potential market for the goods these factories produced. As a result, these governments sought to safeguard their commercial interests in Africa and began sending scouts to the continent to secure treaties from indigenous peoples or their supposed representatives. Similarly, Belgium’s King Leopold II, who aspired to increase his personal wealth by acquiring African territory, hired agents to lay claim to vast tracts of land in central Africa. To protect Germany’s commercial interests, German Chancellor Otto Von Bismarck, who was otherwise uninterested in Africa, felt compelled to stake claims to African land.

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A biological community in Yellowstone National Park most likely includes
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HEYY I NEED HELP WITH AN AP CALC ASSIGNMENT ASAP
Vladimir79 [104]

CALCULATOR PART

1. The area of R + S is unsigned, meaning you want to find

\displaystyle\int_a^b\left|f(x)-g(x)\right|\,\mathrm dx

where [a,b] is the interval between the leftmost and rightmost intersections of f and g.

First use your calculator to find these intersections:

\cos x=\dfrac{x+1}3\implies x\approx-3.64,x\approx-1.86,x\approx0.889

so that a=-3.64 and b=0.889. Now compute the integral using your calculator:

\displaystyle\int_a^b\left|f(x)-g(x)|\,\mathrm dx\approx1.662

2. The volume, using the washer method, is given by the integral

\displaystyle\pi\int_{-1.86}^{0.889}(|2-g(x)|^2-|2-f(x)|^2)\,\mathrm dx\approx12.078

3. A circle of radius r has area \pi r^2; a semicircle with the same radius thus has area \frac{\pi r^2}2. Each cross section of this solid is a semicircle whose diameter is the vertical distance between f(x) and g(x), or |f(x)-g(x)|. In terms of the diameter d=2r, the area of each semicircle would be \frac{\pi d^2}8. Then the volume of the solid is

\displaystyle\frac\pi8\int_{-3.64}^{-1.86}|f(x)-g(x)|^2\,\mathrm dx\approx0.0425

NON-CALCULATOR PART

4. The mean value theorem says that for a function F continuous on an interval [a,b] and differentiable on (a,b), there is some c\in(a,b) such that

F'(c)=\dfrac{F(b)-F(a)}{b-a}

If this F happens to be an antiderivative of f, then we end up with

f(c)=\displaystyle\frac1{b-a}\int_a^bf(x)\,\mathrm dx

\cos x is continuous and differentiable everywhere, so the MVT applies. We have F'(x)=f(x)=\cos x, so the MVT tells us there is some c\in[0,\pi such that

\cos c=\dfrac{\sin\pi-\sin0}{\pi-0}=0

That is, the average value of f(x) on [0,\pi] is 0. The MVT says there is some c in the interval such that the function takes on the average value itself; this happens for c=\frac\pi2.

5. This question seems to be incomplete...

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