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mash [69]
3 years ago
5

Which lists the fractions and in order from least to greatest? 9/5,13/8,1 3/4

Mathematics
1 answer:
adoni [48]3 years ago
8 0

Answer:

A

Step-by-step explanation:

13/8, 1 3/4, 9/5

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Simplify the expression. Write your answer as a power.
nekit [7.7K]

Answer:

either 2^8/3^8 or 256/6561

Step-by-step explanation:

looked it up on Symbolab

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3 years ago
Given g(x)=-x-1, find g(-4)
Molodets [167]

Answer:

g(-4)=3

Step-by-step explanation:

g(x)=-x-1

g(-4)=-(-4)-1

g(-4)=4-1

g(-4)=3

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3 years ago
<img src="https://tex.z-dn.net/?f=100%20%5Ctimes%2068%20%5Cdiv%2015" id="TexFormula1" title="100 \times 68 \div 15" alt="100 \ti
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The answer is 453 rounded
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3 years ago
The sum of a rational number and an even integer is rational. A) Always True B) Sometimes True C) Usually True D) Never True
SOVA2 [1]
I think that the sum will always be a rational number
let's prove that


<span>any rational number can be represented as a/b where a and b are integers and b≠0

</span>and an integer is the counting numbers plus their negatives and 0
so like -4,-3,-2,-1,0,1,2,3,4....

<span>so, 2 rational numbers can be represented as

</span>a/b and c/d (where a,b,c,d are all integers and b≠0 and d≠0)

their sum is
a/b+c/d=
ad/bd+bc/bd=
(ad+bc)/bd

1. the numerator and denominator will be integers
2. that the denominator does not equal 0

alright
1.
we started with that they are all integers
ab+bc=?
if we multiply any 2 integers, we get an integer
<span>like 3*4=12 or -3*4=-12 or -3*-4=12, etc.
</span>even 0*4=0, that's an integer
the sum of any 2 integers is an integer
like 4+3=7, 3+(-4)=-1, 3+0=3, etc.
so we have established that the numerator is an integer

now the denominator
that is just a product of 2 integers so it is an integer


<span>2. we originally defined that b≠0 and d≠0 so we're good

</span>therefore, the sum of any 2 rational numbers will always be a rational number <span>is the correct answer.</span>
6 0
3 years ago
Show all relevant steps.
Usimov [2.4K]
The answer is A hope this helps!
7 0
3 years ago
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