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zalisa [80]
3 years ago
10

What are means of 8/3 = 2/n

Mathematics
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

8/3=2/0.75

Step-by-step explanation:

The variable that you are looking for should be the simplified version of the number on the left.

That is where the 2 would come in. Both numbers should be divided by 4 to gat the answer.

8÷4 = 2

3÷4=0.75

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2 Find the area of the shape. Either enter an exact answer in terms of it or use 3.14 for it and enter your answer as a decimal.
Lady bird [3.3K]

Answer:

The problem is not provided so I cannot solve anything.

Step-by-step explanation:

3 0
2 years ago
<img src="https://tex.z-dn.net/?f=%28%20%7Bx%7D%5E%7B3%7D%20%2B%20%20%7Bx%7D%5E%7B2%7D%20%2B%20x%20%2B%202%29%20%5Cdiv%20%20%28%
Roman55 [17]

(x^3+x^2+x+2) divide by (x^2-1)

We use long division to divide

There is no x term in x^2 -1  so we put 0x

                         x   +    1

                        ----------------------------

x^2+0x-1           x^3+ x^2 + x+ 2

                         x^3+0x^2-x

                         -----------------------------(subtract the bottom from top)

                               x^2 +2x + 2

                               x^2 +0x - 1

                          --------------------------------(subtract the bottom from top)

                                      2x + 3

                    -----------------------------------------    

Quotient : x+1

Remainder : 2x+3

                               


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
The capacity of a beaker is 0.1 liter.convert this to milliliters
Anna [14]
0.1 liters = 100 milliliters 
5 0
3 years ago
suppose you are the financial manager of the International Emirates University IEU. And you are requested at the end of the year
Irina18 [472]
So what this is is
many words

assuming year 0 is 2017

so compound first thing till 2020, take out 30000
the remaining is copmpounded til 2022, take out 50000
remaining is compounded for 1 more year and that is equal to 80000

so from 2017 to 2020, that is 5 years
from 2020 to 2022 is 2 years
from 2022 to 2023 is 1 year


work backwards

A=P(r+1)^t
last one
A=80000
P=?
r=0.08
t=1 year

80000=P(1.08)^1
divide both sides by 1.08
I would leave in fraction
20000000/27=P

now that is the remaining after paying 50000, after 2 years of compounding

so
50000+(2000000/27)=P(1.08)^2
solve using math
about
106374=P

now reverse back

5 years
paid 30000

30000+106374=P(1.08)^5
solve using math
92813.526=P
round
$92813.53




put $92813.53 in the fund
8 0
3 years ago
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