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Vesnalui [34]
3 years ago
13

PLEASE HELP!!! I WILL GIVE BRAINLIEST AND 100 POINTS

Mathematics
2 answers:
Mice21 [21]3 years ago
8 0

Answer:

2x - 2y or x -2y or 0pick one out of the three

Fiesta28 [93]3 years ago
5 0

Answer:

B

Step-by-step explanation:

So we have the equation:

4xz-4yz=5

And we want to find the value of 5/4x in terms of x and y.

From our original equation, let's factor out a 4z from the left. So:

4z(x-y)=5

Now, divide both sides by 4z:

x-y=\frac{5}{4z}

Therefore, the expression that represents 5/4z is x-y.

The correct answer is B.

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3 years ago
Need help i dont get it <br> So please help me
aliina [53]
Confusing.....!!!! sorry
5 0
3 years ago
The sum of 2 composite numbers is never a prime number. Explain your answer.
Natasha2012 [34]

Answer:

Step-by-step explanation:

Composite numbers are positive numbers that have factors, This means that they are divisible by numbers other than 1 and itself provided that number is a factor of the composite number. They possess at the bearest minimum level, a divisor other than  1 and itself. They are a natural number that is expressible as the product of two(or more) numbers other than 1 and itself.

For example:

4 is a composite number because its factors are 1, 2 and 4 which have another divisor apart from 1 and itself (4). That divisor is 2.

We all know that prime numbers are numbers that can be only be divided by 1 and itself.

Therefore, the sum of two composite number, for example:

4 + 6 = 10, We can now see that 10 is never a prime number.

5 0
4 years ago
Jeff's net monthly income is $2550. His monthly expense for rent is $625. What percent of his net monthly income is his rent? (R
kakasveta [241]

Answer:

25%

I cannot really describe how I did it but I am pretty sure it is correct.

3 0
4 years ago
Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.
alina1380 [7]

Answer:

The answer is x=a^{-1}cb^{-1}.

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element g\in G there exist another element, denoted by g^{-1} such that gg^{-1}=e, where e stands for the identity element of G.

So, given the equality axb=c we make a left multiplication by a^{-1} and we obtain:

a^{-1}axb =a^{-1}c.

But, a^{-1}axb = exb = xb. Hence, xb = a^{-1}c.

Now, in the equality xb = a^{-1}c we make a right multiplication by b^{-1}, and we obtain

xbb^{-1} = a^{-1}cb^{-1}.

Recall that bb^{-1}=e and xe=x. Therefore,

x=a^{-1}cb^{-1}.

6 0
3 years ago
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