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MaRussiya [10]
3 years ago
14

PLEASE HELP MATH EXPAND LOG

Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

½log3 + ½logx

Step-by-step explanation:

½(log(3x))

½(log3 + logx)

½logx + ½log3

You might be interested in
What is 0.8 and 5/8 and 0.01 in a percentage
vovikov84 [41]

The correct answers are:

__________________________

   1)      " 0.8  =  80 % "  .

__________________________

   2)      " \frac{5}{8}  =  62. 5 % " .

__________________________

   3)     " 0.01  =  1 % " .

__________________________

Explanation:

To solve:

__________________________

Write the numbers:  " 0.8 " ;   " 5/8 " ;  <u><em>and</em></u>:  " 0.01 " ;  as "percentages" .

To write a "number" as a "percentage" :

  • You take the "number" ;
  • Multiply that "number"  by "100"  ;   <em><u>AND</u></em>:
  • add a  "percent symbol";

                 →   that is:  " % " ;

                 → to denote the "percentage" ;  or, "parts per hundred" .

__________________________

Let us start with the number:  " 0.8 " .

→  " 0.8 "  *  100  =  ?

→   Take the decimal point, and move it 2 decimal spaces to the right; since we are "multiplying" by "100" (which has "2 zeros")  ;

→    0.8  * 100 = ?  ;   If we move the decimal space "one space" forward, we have:  "8" .  If we move the decimal space yet "one more space forward" ;

      →  "8."  to:   " 80 " .

→ Then we add the "percent symbol":

     →  " 80 "  ;   to:  " 80 % "

Answer:  " 0.8  =  80 % " .

_________________________

Other method:

_________________________

→ Note that "percentage" means:  "parts per hundred" :

→ Given:  " 0.8 " ;  Convert to a "percentage" :

→  " 0.8  = \frac{8}{10} " ;

        →  " \frac{8}{10}  = \frac{?}{100} " ;

        →   Find the "?" value.

        →  To do so; look at the "denominators" :

                 →  " 10 * (what) = 100 " ?

                 →  Divide each side by "10" ;

                 → " [10 * what ] / 10  = 100 / 10 " ;

                 →  to get:  

                 →  " [what] " =  "10" .        

                 →    So;   " \frac{8}{10}  = \frac{?}{100} " ;

                 →    If  "{10 * 10 = 100}" ;  then, "{8 * 10 = 80}" .

                 →    So;  " {?} " =  " 80 " .

                 →    So;    " \frac{8}{10}  = \frac{80}{100} " ;

                 →    As such:   " 0.8 "    =  " (\frac{80}{100}) " ;

                                                       =   " 80 "   parts per hundred  ;

                                                       =   " 80  % " .

Answer: " 0.8 "   =   80 %  "  .                                    

__________________________

Now, let convert:  "(\frac{5}{8})" ; into a percentage.

→  Multiply by "100" ;

→  "(\frac{5}{8})"   *   100  ;

→     "(\frac{5}{8})    *   (\frac{100}{1})"  ;

→     "(\frac{(5*100)}{(8*1)})"   ;

→   "(\frac{500}{8})"  ;

   =    " { 500 ÷ 8 } "  ;

  =   " 62. 5 "  ;

→  Now, add the "percent symbol" ;

    →    "  62. 5 "   ;    →  62. 5 % " .

Answer:  " \frac{5}{8}  =  62. 5 % " .

__________________________

Now, let us finish with our last "given value":

Convert:  " 0.01 " ;  to a "percentage" :

→  " 0.01 "  *  100  =  ?

→   Take the decimal point, and move it 2 decimal spaces to the right; since we are "multiplying" by "100" (which has "2 zeros")  ;

→    0.01  * 100 = ?  ;   If we move the decimal space "2 spaces forward" , we have:  " 1 " .  

As such:

      →  " 0.01 "  to:   " 1 " .

→ Then we add the "percent symbol":

     →  " 1 "  ;   to:  " 1 % "  .

Answer:   " 0.01  =  1 % " .

__________________________

Hope this answer is of help to you.

Best wishes!

__________________________

5 0
3 years ago
A soft drink machine outputs a mean of 2323 ounces per cup. The machine's output is normally distributed with a standard deviati
viktelen [127]

Answer:

Area under the normal curve: 0.6915.

69.15% probability of putting less than 24 ounces in a cup.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 23, \sigma = 2

You have been asked to calculate the probability of putting less than 24 ounces in a cup.

pvalue of Z when X = 24. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{24 - 23}{2}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

Area under the normal curve: 0.6915.

69.15% probability of putting less than 24 ounces in a cup.

5 0
3 years ago
What is the answer?<br> Please help
Bumek [7]

Answer:

100*

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Between which two integers is the positive value of the square root of 55?
saw5 [17]

I believe its between the integers of 7 and 8.

7 0
3 years ago
Read 2 more answers
Limit as x approaches 9 of x^2 -81/sqrt of x - 3
Ipatiy [6.2K]

Answer:

108

Step-by-step explanation:

Limit as x approaches 9 of x^2 -81/sqrt of x - 3

First substitute x into the expression

= 9²-81/√9 - 3

= 81-81/3-3

= 0/0 (indeterminate)

Apply l'hospital rule

= lim x -> 9 d/dx(x²-81)/√x - 3

= lim x -> 9 2x/1/2√x

Substitute x = 9

= 2(9)/1/2√9

=18/1/(2(3)

=18 × 6/1

= 108

Hence the limit of the function is 108

7 0
3 years ago
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