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3 years ago

What is 3.41 (where the .41 is repeating) written as a fraction?Please help!!

Mathematics

1 answer:

Aneli [31]3 years ago

7 0

Answer:

41/99

Step-by-step explanation:

There are two types of non terminating decimals. These are: Simple and Mixed

The one that you wrote up here is Simple, Since 41 is the only number that goes on repeating itself.

And mixed non terminating decimal is like 0.352 whereas 52 keeps repeating itself.

So when you change a non terminating decimal the denominator is always 9. But it depends on the decimal whether it is simple or mixed.

Since the decimal you wrote is simple and 2 digits keep on repeating themselves the denominator will be 99.

And the numerator will be the decimal number that keeps repeating itself without the repeating bar.

Therefore, the answer is 41/99.

Hope it helps ;) ❤❤❤

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The value of f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Given a function f(x)=4-2x+6 $x^{2}$ .

We are told to find out the value of f(a), f(a+h) and [f(a+h)-f(a)]/hwhere h≠0.

Function is like a relationship between two or more variables expressed in equal to form.The value which we entered in the function is known as domain and the value which we get after entering the values is known as codomain or range.

f(a)=4-2a+6 $a^{2}$ (By just putting x=a).

f(a+h)== $4-2(a+h)+6(a+h)^{2}$

=4-2a-2h+6( $a^{2} +h^{2} +2ah$ )

=4-2a-2h+6 $a^{2} +6h^{2} +12ah$

= $6a^{2} +6h^{2}-2a-2h+12ah$

[f(a+h)-f(a)]/h=[ $6a^{2} +6h^{2}-2a-2h+12ah$ -(4-2a+6 $a^{2}$ )]/h

= $(6a^{2} +6h^{2} -2a-2h+12ah)/h$

= $(6h^{2} -2h+12ah)/h$

=6h+12a-2.

Hence the value of function f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Learn more about function at brainly.com/question/10439235

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