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mixas84 [53]
3 years ago
11

Justify why a/b x b/c x c/d x d/ e is equal to a/e when b,c,d, and e are not zero?

Mathematics
1 answer:
Nuetrik [128]3 years ago
5 0
When you multiply fractions, you multiply across, right? so, for example 5/7 times 4/9 will give you 20/63. So, if you do this to this problem, you will have abcd/bcde. but now you are able to cancel out the b and c and d. think of it like reducing. 5/10 is 1/2. Why? because if you reduce/factor out a 5, you get a 1 in the numerator and a 2 in the denominator. same thing here; you can factor out a b, c, and d(because they show up in both num. and denom. so what you are left with is a/e. Cheers!
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A geography teacher shares 18 maps equally between 8 students. The number of maps that each student gets lies between what two w
Serga [27]
I believe it is 2 and 3.
8 0
3 years ago
Twenty-seven is<br>% of 60<br>help mee asap​
jok3333 [9.3K]

Answer:  The answer is:  " <u> </u><u>45 </u>  % "  .    

________________________________________________

               →    " Twenty-seven is <u> 45 </u> % of 60. "

________________________________________________

Step-by-step explanation:

________________________________________________

The question asks:

 " 27 is what % {percentage] of 60 " ?  ;

________________________

So:  " 27 =  (n/100) * 60 " ;  Solve for "n" ;

________________________________________________

Method 1:

________________________________________________

  →   (n/100) * 60 = 27 ;

Divide each side by 60 :

 →   [ (n/100)  * 60 ] / 60 = 27 /60 ;

to get:

 →    (n/100) = 27/60 ;

Now:  Cross-factor multiply:

 →  60n = (27)*(100) ;

to get:

 → 60n = 2700 ;

Divide each side by "60" ;

→  60n = 2700/ 60 ;

to get:  n = 45 ;

________________________

 →  The answer is:  45 % .    

   →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Method 2:

________________________________________________

The question asks:

 " 27 is what % {percentage] of 60 " ?

________________________

To solve this problem:

Rephrase this question as:

________________________

" 27 is 60% of what number ? "

 →  The answer will be the same!

________________________

→  27 = (60/100)* n ;   Solve for "n" ;

Multiply each side of the equation by "100" ; to eliminate the fraction:

→  100 * 27 = 100 * [ (60/100)* n ] ;

 to get:

   →   2700 = 60n ;

↔  60n = 2700 ;

Divide Each Side of the equation by "60" ;

    →   60n/60 = 2700 / 60 ;

to get:  n = 45 ;

________________________________________________

→  The answer is:  45 % .    

       →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Method 2:  Variant 1 of 2:

________________________________________________

When we have:

→  27 = (60/100)* n ;   Solve for "n" ;

________________________

Note that:  "(60/100) = (60÷ 100) = (6 ÷ 10)" ;   since:  in "(60/100)" ;  the "zero" from the "<u>numerator</u>" cancels out;  <u>And</u>:  the "last zero" in "100" — from the "<u>denominator</u>" cancels out;  since we are dividing "each side" of the fraction by "10" ;

  →   "(60÷10) / (600÷10)"  =  " 6/10 " ;  

  →   " (6/10)" ; that is;  "six-tenths"} ;  

  →     can be represented by:  " 0.6 " ;

  →  {by convention;  but specifically, here is the explanation} — as follows:

________________________

  →   "(6/10)" =  " (6 ÷ 10) " ;  

<u>Note</u>:  When dividing a number by "10" ;  we take the original number; and move the decimal point to the left; & then we rewrite that number as the "answer".  

<u>Note</u>:  When multiplying or dividing by a positive, non-zero integer factor of "10" that has at least 1 (one) "zero" after that particular factor of "10".  We can get the answer by taking the original number & moving the decimal point the number of spaces as designated by the number of zeros following the particular [aforementioned factor of "10".].

We move the decimal point to the right if we are multiplying;  and to the left  if we are dividing.  In this case, <u>we are dividing</u> "6" by "10 " :

 →  " 6   ÷  10  =  ? " ;  

 →  " 6.  ÷ 10 =  ? " ;

   We take the: " 6. " ;  and move the decimal point "<u>one space backward [i.e. "to the left</u>"];  since we are <u>dividing by "</u><u>10</u><u>"</u> ;

 →  to get:  " .6 " ;  & we rewrite this value as "0.6" in a rewritten equation:

________________________

So; we take our equation:

→  27 = (60/100)*n ;  And rewrite—substituting "0.6" for

"(60/100)"— as follows:

________________________

→  27 = (0.6)n ;  ↔ (0.6) n = 27 ;

Multiply each side of the equation by "10" ; to eliminate the decimal:

   →  10 * [ (0.6)n ]  = 27 * 10 ;

to get:

  →  6n = 270 ;

Divide each side of the equation by "6" ; to isolate "n" on one side of the equation; & to solve for "n" ;

 →  6n / 6  =  270 / 6 ;

to get:   n = 45 ;

________________________________________________

→  The answer is:  45 % .    

      →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Method 2 (variant 2 of 2):

________________________________________________

We have the equation:  27 = (60/100)* n ;   Solve for "n" ;

________________________

<u>Note</u>:  From Method 2 (variant) 1 of 2— see above):

________________________

<u>Note</u>:  Refer to the point at which we have:

________________________

→   " {  (60÷10) / (600÷10)"  =  " (6/10) " ;  that is;  "six-tenths"} ;

________________________

Note that the fraction— "(6/10)" ;  can be further simplified:

→  "(6/10)" =  "(6÷2) / (10÷2)" = "(3/5)" ;

Now, we can rewrite the equation;

→ We replace "(60/100)" ;  with:  "(3/5)" :

    →  27 = (3/5)* n ;   Solve for "n" ;

↔ (3/5)* n = 27 ;  

↔    (3n/5) = 27 ;

Multiply Each Side of the equation by "5" ;

→  5* (3n/5) = 27 * 5 ;  

to get:

→   3n = 135 ;

Divide Each side of the equation by "3" ;  to isolate "n" on one side of the equation;  & to solve for "n" ;

→  3n / 3 = 135 / 3  ;

to get:   n = 45 ;

________________________________________________

 →  The answer is:  45 % .    

       →  " Twenty-seven is <u>45 %</u> of 60."

________________________________________________

Hope this answer is helpful!

        Wishing you the best in your academic endeavors

           — and within the "Brainly" community!

________________________________________________

7 0
3 years ago
Read 2 more answers
9. The ticket at right has a perimeter of 42 cm.
Liula [17]

Perimeter simply represents the sum of all side lengths of a shape. The length of the missing side of the ticket is 5 cm; the length of gold line on each ticket is 20 cm, and 2 bottles of gold ink are required to draw gold lines on 200 tickets.

I've added the image of the ticket as an attachment.

(a) The missing side length

From the attachment, the 4 unknown side lengths are equal. Represent this side length with L.

So, we have:

Perimeter =2\times 11 + 4 \times L

This gives

2\times11 + 4 \times L = 42

22 + 4 \times L = 42

Collect like terms

4 \times L = 42-22

4 \times L = 20

Divide both sides by 4

L =5cm

(b) The length of the gold lines

There are 4 slant lines and the length of one of the slant lines is 5 cm (as calculated above).

So, the length of the gold line is:

Gold = 4 \times L

Gold = 4 \times 5cm

Gold = 20cm

(c) The number of gold ink bottles.

n = 200 --- number of tickets

The length of all gold line in the 200 tickets is:

Length = 200 \times Gold

Length = 200 \times 20cm

Length = 4000 cm

Length = 4000 \times 0.01m ---- convert to meters

Length = 40m

Given that:

Bottle = 20m --- 1 bottle for 20 m

The number of bottles (n) is:

n = \frac{Length}{Bottles}

n = \frac{40m}{20m}

n = 2

Hence, 2 bottles of gold ink are enough.

Read more about perimeters at:

brainly.com/question/6465134

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Step-by-step explanation:

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Ls Figure A'B'C'D' a translation of Figure ABCD? (Click in the picture to see the rest
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