Advanced Placement (AP), 29.07.2019 05:30 10040816

In the early 1970s first used the term "information skills"

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The United States National Forum on Information Literacy defines information literacy as "... the ability to know when there is a need for information, to be able to identify, locate, evaluate, and effectively use that information for the issue or problem at hand."[1][2] The American Library Association defines "information literacy" as a set of abilities requiring individuals to "recognize when information is needed and have the ability to locate, evaluate, and use effectively the needed information.[3] Other definitions incorporate aspects of "skepticism, judgement, free thinking, questioning, and understanding..."[4] or incorporate competencies that an informed citizen of an information society ought to possess to participate intelligently and actively in that society.[5]

Answer from: sarinaneedshelp01

Paul Zurkowski

Explanation:

Paul Zurkowski was the person that created the term information skills in 1974. He used the term for the first time in a document presented to the US National Commission on Libraries and Information Science to refer to techniques that are learned by information literates when they find information solutions to problems by using information resources. According to this, the answer is that in the early 1970s, Paul Zurkowski first used the term "information skills".

Answer from: Quest

a.10 children i think

Answer from: Quest

See diagram.we use washers to get the area and we will be integrating with respect to xouter radius is the distance between y = e^x and y = -1 $r_{\text{out} } = e^x ) = e^x + 1$ inner radius of washer is the distance between y = 1 and y = -1 $r_{\text{in} } = 1 - (-1) = 2$ the area of one cross-section of a washer is $\begin{aligned} a(x) & = \pi(r_{\text{out} } )^2 - \pi(r_{\text{in} } )^2 \\ & = \pi(e^x + 1 )^2 - \pi(2)^2 \\ & = \pi \big[(e^x + 1 )^2 - 4 \big] \end{aligned}$ solid lies between x = 0 and x = 2. the volume is $\begin{aligned} v & = \int_0^2 a(x) dx \\ & = \int_0^2 \pi \big[(e^x + 1 )^2 - 4 \big] dx \end{aligned}$ if you need this expanded, then $\begin{aligned} v & = \int_0^2 \pi \left(e^{2x} + 2e^x + 1 - 4 \right) dx \\ & = \int_0^2 \pi \left(e^{2x} + 2e^x -3 \right) dx \end{aligned}$ the question doesn't say to evaluate, so we're done.

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