Answer:
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The mass of one liter of water is about one ____
1 answer:
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Answer:
35/37
Step-by-step explanation:
In this question, we are to calculate the ratio that represents the cosine of angle O
Firstly, please check attachment for diagram of the triangle.
Mathematically, the trigonometric term Cos means the ratio of the length of the adjacent to that of the hypotenuse
The hypotenuse is the longest length which is 37 which represents the length ON
The side adjacent to angle O is the side OP which is 35
Thus, the ratio representing the cosine of angle O is 35/37
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} ;
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→ We write this value, as a fraction, as: "12/99" ;
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→Explanation:
__________________________________________
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;
_____________________________________________________
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.
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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 ;
__________________________________________________
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: .
_________________________________________
______________________________
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:
_________________________________________