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stira [4]
3 years ago
6

Which of these has a value greater than 3/4? 1/2, 1/4, 1/5

Mathematics
2 answers:
Tju [1.3M]3 years ago
8 0
None of these three fractions has a value greater than 3/4. If you solve these fractions to find their respective decimals, we get

3/4 = 0.75
1/2 = 0.5
1/4 = 0.25
1/5 = 0.2

As we can see, the decimal value of 3/4 (0.75) is the greatest on this list, therefore none of the other fractions has a value greater than 3/4.
Hope that helped =)
NeTakaya3 years ago
7 0
I really think none of them are 3/4 is almost a whole 1/2 is half of a whole 1/4 and 1/5 is close to 0
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Answer:

2/3, 5/7, 16/12

From least to greatest!

Smallest = 2/3, Second = 5/7, Largest = 16/12

Hope this helped :-)

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3 years ago
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erik [133]

Answer:

explanation below

Step-by-step explanation:

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3 0
3 years ago
Is that correct ????
Travka [436]
Yes I believe it is :)
7 0
3 years ago
Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal
lisabon 2012 [21]

Given the following functions below,

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}

Factorising the denominators of both functions,

Factorising the denominator of f(x),

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}

Factorising the denominator of g(x),

\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}

Multiplying both functions,

undefined

4 0
1 year ago
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