The figure below is a rectangle. What is the value of x?
2 answers:
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The answer is: " 
" .
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Given: 0.1212121212..... repeating ; write that value as a fraction;
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In other words; we are given: "0.1212121212..... repeating infinitely" ;
→ that is to say; "0.12 ... ; {the "12" decimal portion repeats infinitely} ;
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→ We write this value, as a fraction, as: "12/99" ;
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→Explanation:
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Note: "0.99999999...... repeating infinitely; = "1" .
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→Since:
Let us say that we have:
"x = 0.999999 ; repeating infinitely;
In order words, let us say we have: "x = 0.9.... ; the "9" decimal repeats infinitely;
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Then "10x" ; (that is: "10" multlipled by "x"; or "10*x" or "10x" ); is equal to:
"10" * (0.999999.....) = 9.99999999...... (the "9" decimal repeats infinitely);
in other words: 10x = 9.99999999....
Divide each side by "10" ;
to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;
But if you have: "10x = 10" ; divide each side of the equation by "10" ;
you get: "x = 1" .
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Also, if "x = 0.9999...(repeating infinitely);
then: 10x = 9.99999.
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10x = 9.999999999999999......
− x = 0.999999999999999.......
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9x = 9.00000000000000000000.....
→ 9x = 9 ;
Divide each side of the equation; to get;
9x /9 = 9/ 9 ; to get: x = 1 ; and we have: x = 0.9999.... ; so
x= 0.99999.... = 1 ;
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So, if the numbers "12" is repeating, we divie "12" by "99" ;
that is; we divide "12" by "two 9's" ; since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
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So; 0.12121212.....(the "12" is the decimal that repeats infinitely);
= 12/99 ; which can be simplified;
Divide each side (both the numerator AND the denominator); by "3" ;
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" 12/99 " = "(12÷3) / (99÷3) = 4/33 " .
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The answer is: .
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______________________________
Given: 0.1212121212..... repeating ; write that value as a fraction;
______________________________________________________
In other words; we are given: "0.1212121212..... repeating infinitely" ;
→ that is to say; "0.12 ... ; {the "12" decimal portion repeats infinitely} ;
_______________________________________________________
→ We write this value, as a fraction, as: "12/99" ;
__________________________________________
→Explanation:
__________________________________________
Note: "0.99999999...... repeating infinitely; = "1" .
_________________________________________
→Since:
Let us say that we have:
"x = 0.999999 ; repeating infinitely;
In order words, let us say we have: "x = 0.9.... ; the "9" decimal repeats infinitely;
_____________________________________________________
Then "10x" ; (that is: "10" multlipled by "x"; or "10*x" or "10x" ); is equal to:
"10" * (0.999999.....) = 9.99999999...... (the "9" decimal repeats infinitely);
in other words: 10x = 9.99999999....
Divide each side by "10" ;
to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;
But if you have: "10x = 10" ; divide each side of the equation by "10" ;
you get: "x = 1" .
____________________
Also, if "x = 0.9999...(repeating infinitely);
then: 10x = 9.99999.
_______________________________________________
10x = 9.999999999999999......
− x = 0.999999999999999.......
_____________________________________
9x = 9.00000000000000000000.....
→ 9x = 9 ;
Divide each side of the equation; to get;
9x /9 = 9/ 9 ; to get: x = 1 ; and we have: x = 0.9999.... ; so
x= 0.99999.... = 1 ;
__________________________________________________
So, if the numbers "12" is repeating, we divie "12" by "99" ;
that is; we divide "12" by "two 9's" ; since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
________________________________________
So; 0.12121212.....(the "12" is the decimal that repeats infinitely);
= 12/99 ; which can be simplified;
Divide each side (both the numerator AND the denominator); by "3" ;
_________________________________________
" 12/99 " = "(12÷3) / (99÷3) = 4/33 " .
_________________________________________
The answer is:
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Remark
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer
This question likely should be done before the other one. What you are trying to do is give C a value. So you need to remember that C is always part of an indefinite integral.
y =
y = sin(x) - cos(x) + C
y(π) = sin(π) - cos(π) + C = 0
y(π) = 0 -(-1) + C = 0
y(π) = 1 + C = 0
C = - 1
y = sin(x) - cos(x) - 1 <<<<< Answer
Problem Two
Remember that
y( - e^3 ) = ln(|x|) + C = 0
y(-e^3) = ln(|-e^3|) + C = 0
y(-e^3) = 3 + C = 0
3 + C = 0
C = - 3
y = ln(|x|) - 3 <<<< Answer