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Sophie [7]
2 years ago
8

Find the value of f(-8).

Mathematics
1 answer:
fredd [130]2 years ago
5 0

Answer:

-8f

Step-by-step explanation:

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1/6 divided by 2

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1/6 * 1/2 = 1/12

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How many centimeters(cm) of rain fell in this region during the course of the year​
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Suppose a shoe factory produces both low-grade and high-grade shoes. The factory produces at least twice as many low-grade as hi
umka2103 [35]

Answer:

The factory should produce 166 pairs of high-grade shoes and 364 pairs of low-grade shoes for maximum profit

Step-by-step explanation:

The given parameters for the shoe production are;

The number of low grade shoes the factory produces ≥ 2 × The number of high-grade shoes produced by the factory

The maximum number of shoes the factory can produce = 500 pairs of shoes

The number of high-grade shoes the dealer calls for daily ≥ 100 pairs

The profit made per pair of high-grade shoed = Birr 2.00

The profit made per of low-grade shoes = Birr 1.00

Let 'H', represent the number of high grade shoes the factory produces and let 'L' represent he number of low-grade shoes the factory produces, we have;

L ≥ 2·H...(1)

L + H ≤ 500...(2)

H ≥ 100...(3)

Total profit, P = 2·H + L

From inequalities (1) and (2), we have;

3·H ≤ 500

H ≤ 500/3 ≈ 166

The maximum number of high-grade shoes that can be produced, H ≤ 166

Therefore, for maximum profit, the factory should produce the maximum number of high-grade shoe pairs, H = 166 pairs

The number of pairs of low grade shoes the factory should produce, L = 500 - 166 = 334 pairs

The maximum profit, P = 2 × 166 + 1 × 364 = 696

6 1
3 years ago
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Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
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