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Darya [45]
3 years ago
10

Define reductionism i m so boredd​

Mathematics
2 answers:
mihalych1998 [28]3 years ago
8 0

Answer:

<h3>Reductionism, in philosophy, a view that asserts that entities of a given kind are identical to, or are collections or combinations of, entities of another (often simpler or more basic) kind or that expressions denoting such entities are definable in terms of expressions denoting other entities.</h3>

<h2><u>hope </u><u>it</u><u> </u><u>helps</u></h2>

<h2><u>stay </u><u>safe </u><u>healthy</u><u> </u><u>and</u><u> </u><u>happy</u><u>.</u></h2>
Sidana [21]3 years ago
4 0

Step-by-step explanation:

Reductionism is a theory in psychology centered on reducing complex phenomena into their most basic parts. It is often contrasted with holism, which is focused on looking at things as a whole.

The purpose of reductionism is to simplify psychological events and processes by looking at their smallest elements, thus “reducing” something rather complex into its most simple.

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Which equation is an identity a; -3y 3=-3y-6b; 3w 8-w=4w-2(w-4)c;9-(2v 3)=-2v-6d; 7m-5=8m 7-m?
Marta_Voda [28]
3w + 8 - w = 4w - 2(w - 4)
2w + 8 = 4w - 2w + 8
2w + 8 = 2w + 8
0 = 0

Therefore, option b is an identity.
4 0
3 years ago
Let A and B be subsets of a universal set U and suppose n(U)=400, n(A)=135, n(B)=80, and n(A n B)=60. Find the number of element
Alexxandr [17]
Use A' for A complement, etc. and n = intersection &cap;.
n(A' n B')
=n(A u B)'   by the de Morgan's law
=nU-n(nA+nB-n(AnB))     n(A u B) = nA + nB - n(AnB)  and nX'=nU-nX
=400-(135+80-60)
=400-155
=245
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3 years ago
What is the 3d version of a oval
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7 0
4 years ago
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John is making a recipe that call for 1/4 cup of flour for every 1/2 cup of water. How much water will he need if he uses 1 1/2
irakobra [83]
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7 0
3 years ago
Rationalise the denominator of the following​
Georgia [21]

Answer:

2 + √3

Step-by-step explanation:

Multiply the expression by \dfrac{\sqrt{3} +1}{\sqrt{3} +1} :

\implies \dfrac{\sqrt{3} +1}{\sqrt{3} -1}\times\dfrac{\sqrt{3} +1}{\sqrt{3} +1}=\dfrac{(\sqrt{3} +1)^2}{3+\sqrt{3}-\sqrt{3}-1}=\dfrac{(\sqrt{3} +1)^2}{2}

\implies \dfrac{(\sqrt{3} +1)^2}{2}=\dfrac{3+2\sqrt{3}+1 }{2}=\dfrac{4+2\sqrt{3} }{2}=2+\sqrt{3}

4 0
2 years ago
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