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Alex
3 years ago
5

Simplify completely quantity 6 x squared minus 54 plus 84 over quantity 8 x squared minus 40 x plus 48 divided by quantity x squ

ared plus x minus 56 over quantity 2 x squared plus 12 x minus 32
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

After solving the given problem, I’ve been able to get 3(x-2) / 2(x-3) as the simplified form of the given equation. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

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Look at the box-and-whisker plot. What is the measure of the first quartile (Q1)?
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Please provide the Box and whisker plot so that we can answer the question.
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Over a 5-year period.
kaheart [24]

Answer:

Year 0 900$

Year 1 1000$

year 2 1080$

Year 3 61040$- (interest comes in)

Year 4 110000$

Year 5 152000$

Step-by-step explanation:

Make sure you put that interest comes in or you will be marked wrong.

You're welcome :D GoodLuck have a good day

7 0
3 years ago
Need assistance with ratios
jeyben [28]

Answer:

1. 32

4:4 is the same as 1:1 meaning each one gets the same

2. 40

put 1/7 equal to 5/x and cross multiply (1x=35)(35 male workers, so add 35 to 5 female workers)

3. 8

Multiply two by 4 since an hour is 15×4

4. 188

put the ratio 7/330 to 4/x (7x=1320)(x=188.57) since you can use part of a nail, round down to 188.

5. 5 1/2 pounds

multiply 1/4 by 22 which is 22/4. that is equal to 5.5

5 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
How do u spell 5,165,874 in word form
Novay_Z [31]
Five million one hundred sixty five thousand eight hundred seventy four.
4 0
3 years ago
Read 2 more answers
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