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ankoles [38]
3 years ago
12

Can someone please help me with this question:)

Mathematics
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

x = 12 and y = 24 Can I get brainiiest

Step-by-step explanation:

We know (3x + 34) and (2x + 46) are vertical to each other so they equal the same measure. We will solve it like this.

3x + 34 = 2x + 46 (subtract 34 on each side)

3x = 2x + 12 (subtract 2x on both sides)

x = 12

Now plug in

3(12) + 34 = 2(12) + 46

36 + 34 = 24 + 46

70 = 70

Now all you have to do is do the other pair of angle

4y + 14 + 70 = 180 (we are using what we already know about 70)

(collect like terms)

4y + 84 = 180 (subtract 84 on both sides)

4y = 76 (divide by 4)

y = 24

(plug in)

4(24) + 14 = 6(24) - 34

96 + 14 = 144 - 34

110 = 110

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Answer:x+y=8

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Determine,in each of the following cases, whether the described system is or not a group. Explain your answers. Determine what i
zheka24 [161]

Answer:

(a) Not a group

(b) Not a group

(c) Abelian group

Step-by-step explanation:

<em>In order for a system <G,*> to be a group, the following must be satisfied </em>

<em> (1) The binary operation is associative, i.e., (a*b)*c = a*(b*c) for all a,b,c in G </em>

<em>(2) There is an identity element, i.e., there is an element e such that a*e = e*a = a for all a in G </em>

<em> (3) For each a in G, there is an inverse, i.e, another element a' in G such that a*a' = a'*a = e (the identity) </em>

<em> </em>

If in addition the operation * is commutative (a*b = b*a for every a,b in G), then the group is said to be Abelian

(a)  

The system <G,*> is not a group since there are no identity.  

To see this, suppose there is an element e such that  

a*e = a

then  

a-e = a which implies e=0

It is easy to see that 0 cannot be an identity.

For example  

2*0 = 2-0 = 2

Whereas

0*2 = 0-2 = -2

So 2*0 is not equal to 0*2

(b)

The system <G,*> is not a group either.

If A is a matrix 2x2 and the determinant of A det(A)=0, then the inverse of A does not exist.

(c)

The table of the operation G is showed in the attachment.

It is evident that this system is isomorphic under the identity map, to the cyclic group

\mathbb{Z}_{5}

the system formed by the subset of Z, {0,1,2,3,4} with the operation of addition module 5, which is an Abelian cyclic group

We conclude that the system <G,*> is Abelian.

Attachment: Table for the operation * in (c)

4 0
3 years ago
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Answer:

BDC 37°

CDE 143°

FDE 37°

FDA 90°

with a hope, this helps you!

3 0
3 years ago
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Pavlova-9 [17]

Answer:

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<u>Step-by-step explanation:</u>

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