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morpeh [17]
3 years ago
6

WILL GIVE BRAINLIST

Mathematics
1 answer:
nignag [31]3 years ago
4 0

Answer:

8.333... Yards.

Step-by-step explanation:

To convert feet to yards, divide the feet by 3.

\frac{25}{3} = \frac{25}{3}/8.333...

To check work

Multiply 8.333... by 3.

8.333 · 3 = 24.999...

It Is Not Exactly 25 Because The 3 Is Infinite But It Is Enough To Prove Your Answer.

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Factor completely
svet-max [94.6K]

Answer:

B.) 2(5x-4)(x+1)

Step-by-step explanation:

10x^2+2x-8

= 2(5x^2 + x - 4)

= 2(5x^2 + 5x - 4x - 4)

= 2[5x(x + 1) - 4(x + 1)]

= 2(5x-4)(x+1)

8 0
3 years ago
Read 2 more answers
Do these look right?
bezimeni [28]

Answer:

yes it look right like the one you have to do 2x+5=39

5 0
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
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
Solve the following quadratic equation for all values of xx in simplest form. 3(x+1)²-15=-12​
suter [353]

Answer:

x = 0, -2

Step-by-step explanation:

3 0
2 years ago
If Leslie travels 1/3 of a mile in 5/6 of an hour and maintains a constant speed, what is her distance per hour?
hammer [34]
D = S × T
S = D ÷ T
T =D ÷S

sorry thats all i can do
6 0
3 years ago
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