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Nat2105 [25]
3 years ago
11

Which is the best estimate for (6.3x10^-2)(9.9x10^-3) written in scientific notation?

Mathematics
2 answers:
ozzi3 years ago
8 0
It should be 6.237*10^-4
IRISSAK [1]3 years ago
5 0

Answer:

<h3>6 \times 10^{-4}.</h3>

Step-by-step explanation:

We are given expression (6.3 \times 10^{-2})(9.9\times 10^{-3}).

In order to multiply above given expression, we need to multiply 6.3 by 9.9 first.

On multiplying 6.3 by 9.9 , we get 62.37.

Now we would multiply  10^{-2} \ by \ 10^{-3}.

Therefore,

10^{-2} \times 10^{-3}= 10^{-2-3}= 10^{-2-3} = 10^{-5}.  

Therefore,

(6.3 \times 10^{-2})(9.9\times 10^{-3}) =62.37 \times 10^{-5}

Again 62.37 \times 10^{-5} could be written as 6.237 \times 10^{-4}.

Now, 6.237 \times 10^{-4} could be round to 6 \times 10^{-4}.

Therefore, correct answer is  6 \times 10^{-4}.

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Answer:

y=0.3x-2

Step-by-step explanation:

first find the gradient

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m=0-(-2)/6-0

m=0.3

then use this equation

y-y1=m(x-x1)

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4 0
3 years ago
Find the common fraction equivalent to 0.12
jonny [76]
The answer is:  " \frac{4}{33} "  .
______________________________
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:  \frac{4}{33}  .
_________________________________________
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Step-by-step explanation:

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