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Sav [38]
3 years ago
11

Peyton has 2 pizzas.each pizza is cut into 10 equal slices.she and her friends eat 14 slices. what part of the pizzas did they e

at? write your awnser as a decimal
Mathematics
1 answer:
WINSTONCH [101]3 years ago
8 0
Since Peyton has 2 pizzas and each is cut into 10 equal slices that would mean 2x10 which is 20 and then it says she and her friends eat 14 slices which means take away 14 minus 20 which is 6 so the decimal for only 6 would be 0.6 which is the answer. Hope this helped.
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Which line has a constant proportionality of 4
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Line A

Step-by-step explanation:

Line A has a constant of proportionality (slope) of 4.

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All the students in the six grade either purchase their lunch or brought their lunch from home on Monday 24% of the students pur
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3 years ago
Determine the numerical length of JK.​
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Answer:

30

So the two sectors add to the full length:

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5 0
3 years ago
Read 2 more answers
(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
Dahasolnce [82]

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.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

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}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

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

7 0
3 years ago
a drawer contains 10 blue pens and 10 red pens Without looking Msr. Stanton takes 1 pen uses it and puts it back
Rudiy27
Answer:
1/2 if it's a probability question of getting one blue pen But there's also very little information on the question
Step-by-step explanation:

4 0
2 years ago
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