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ivolga24 [154]
3 years ago
13

I really need this pls helppp

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
5 0
I’ll try to help but there’s nothing to help with hahaha :)
Morgarella [4.7K]3 years ago
3 0

Answer:

Help with what?? I'll try

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THE THING ON THE TOP IS y=4/3x-4<br> Y=2/3x-6
noname [10]

Answer:

the graphs are difficult to read. The answer is whichever graph has the point (-3,-8) I think option 2.

Step-by-step explanation:

Find where the two equations intersect.

Two equations intersect where they are equal.

(4/3)x-4 = (2/3)x-6

(4/3)x-(2/3)x - 4 = - 6 (subtracted (2/3)x from both sides)

(2/3)x -4 = -6

(2/3)x = -2 (added 4 to both sides)

2x = - 6 (multiplied both sides by 3)

x = - 3 (divided both sides by 2)

At this point we know x is negative so the answer cannot be options 1 or 3.

Let's solve for y plugging x into any of the equations.

y = (2/3)(-3) -6 = -2 - 6 = -8

y = (4/3)(-3) -4 = -4 -4 = -8

So the equations intersect at (x,y)=(-3,-8)

8 0
3 years ago
Image down below thx:)
user100 [1]

Answer:

(1,2)

Step-by-step explanation:

You go over 1 and then up 2 and that is where Q is.

5 0
3 years ago
A glider begins its descent 3/5​​ mile above the ground. After 30 minutes, it is 9/20​​ mile above the ground. If it continues t
Andreas93 [3]

Answer:

I dont know

Step-by-step explanation:

I was looking for the same answer to this question☠

8 0
2 years ago
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
Victor draws one side of equilateral ∆PQR on the coordinate plane at points P (–9, –2) and Q (–2, –2). What are the two possible
Elenna [48]

Answer:

the vertices of point R could be (-2,5) or (-2,9)

each point is 7 units away from Q since there are 7 units between P and Q

Step-by-step explanation:

6 0
3 years ago
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