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xxTIMURxx [149]
3 years ago
9

If you have 2 squares with the same side lengths are they still considerd scaled copies of each other?

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
7 0
Yes as long as one side on a square is the same on another they are the copies of each other
You might be interested in
(2x^2+x+3)/(x-2)with remainder
Leni [432]

Answer:

2x+5 r. 13

Step-by-step explanation:

So using long division, you can solve for the quotient and the remainder.

Please look at the attached for the solution.

Step 1: need to make sure that you right the terms in descending order. (If there are missing terms in between, you need to fill them out with a zero so you won't have a problem with spacing)

Step 2: Divide the highest term in the dividend, by the highest term in the divisor.

Step 3: Multiply your result with the divisor and and write it below the dividend, aligning it with its matched term.

Step 4: Subtract and bring down the next term.

Repeat the steps until you cannot divide any further. If you have left-overs that is your remained.

3 0
3 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
Evaluate the expression<br> 6a - 2b when a = 5, b=7
melomori [17]

6a - 2b

= 6(5) - 2(7)

= 30 - 14

= 16 answer

6 0
3 years ago
Read 2 more answers
-2x &lt; -6
Soloha48 [4]

Answer:

x > 3

Step-by-step explanation:

This is based on the concept 'when a negative number is multiplied to both sides of the inequality, the sign gets reversed. For example, x > - 2 when it is multiplied by - 1, it becomes - x < 2.'

*only when the number is -ve.

Similarly, in this question, multiply both sides by - 1.

2x > 6

x > 6/2

x > 3

For more : take - 5 > - 2, as you know

Note that 5 is actually greater than 2. To make this true we change signs

8 0
3 years ago
How do I do this help due in tomorrow
Mkey [24]

Answer:

The top one is 135 cm^2

Step-by-step explanation:

To find the area, all you need to do is multiply the 3 different measurements. For the one in the top corner we need to multiply the length by the width by the height. The work is done below for you on that one.

A = l * w * h

A = 15 * 3 * 3

A = 135 cm^2

4 0
3 years ago
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