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Phantasy [73]
2 years ago
7

a storage jug was 4/5 full of water when did get indicated 72 what was the full capacity of the storage jug​

Mathematics
1 answer:
IRISSAK [1]2 years ago
6 0

Answer:

<em><u>9</u></em><em><u>0</u></em><em><u>. </u></em><em><u>(</u></em><em><u> </u></em><em><u>was </u></em><em><u>the </u></em><em><u>full </u></em><em><u>capacity</u></em><em><u> of</u></em><em><u> </u></em><em><u>jug</u></em><em><u>)</u></em>

<em><u>to </u></em><em><u>solve </u></em><em><u>this</u></em><em><u> </u></em>

<em><u>firstly</u></em><em><u> </u></em><em><u>let </u></em><em><u>the </u></em><em><u>full </u></em><em><u>capacity</u></em><em><u> </u></em><em><u>of </u></em><em><u>jug </u></em><em><u> </u></em>

<em><u> </u></em><em><u> </u></em>

<em><u>be</u></em><em><u>. </u></em><em><u>x</u></em>

<em><u>so,</u></em><em><u> </u></em>

<em><u>4</u></em><em><u>/</u></em><em><u>5</u></em><em><u>*</u></em><em><u>x </u></em><em><u>=</u></em><em><u> </u></em><em><u>7</u></em><em><u>2</u></em>

<em><u>x </u></em><em><u>=</u></em><em><u> </u></em><em><u>7</u></em><em><u>2</u></em><em><u>*</u></em><em><u>5</u></em><em><u>/</u></em><em><u>4</u></em>

<em><u>x </u></em><em><u>=</u></em><em><u> </u></em><em><u>9</u></em><em><u>0</u></em>

<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em>

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Step-by-step explanation:

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2 years ago
(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho
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Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

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Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

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b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

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