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mihalych1998 [28]
3 years ago
9

Which polynomial identity will prove that 19 = 27 − 8? Difference of Squares Difference of Cubes Sum of Cubes Square of a Binomi

al
Mathematics
2 answers:
Lyrx [107]3 years ago
6 0
The correct answer is:  [B]:  "Difference of Cubes". 
<span>__________________________________________________________

Explanation:
__________________________________________________________
Note that the equation/identity for the "difference of cubes" is expressed as:
___________________________________________________</span><span>_______

        " a</span>³ − b³ = (a − b)(a² + ab + b²) "  ; 
<span>__________________________________________________________

Note the given equation:  " 19 = 27 </span>− 8 " ;   →  (which is true). 
__________________________________________________________

       The "right hand side" of this equation:

            →  " 27 − 8 " ;  contains two numbers:  
         
            →  "27" and "8" ;  both of which are "cubes" ; 

            → that is:  ∛27 = 3 ;   ↔  3³ = 3 * 3 * 3 = 9 * 3 = 27 ;  <u><em>and</em></u>:

                               ∛ 8  = 2 ;  ↔  2³ = 2 * 2 * 2 = 4 * 2 = 8 ; 

            → <u>AND:</u>  "8" is being <u>SUBTRACTED</u> from "27" ; 

            →  (hence, the "difference of squares" polynomial identity); 

So:  given:     " 19 = 27  − 8 " ; 

→  Rewrite as:

        " 19 =  3³  − 2³ " ; 
_______________________________________________________
Now, consider the identity equation for the "difference of squares":

                          →   " a³ − b³ = (a − b)(a² + ab + b²) "  ; 
_______________________________________________________

Take:  " 19  =  3³  − 2³ " ; 

and rewrite as:  

→  3³  −  2³  = 19 ; 

So:   (a³ − b³) = 3³  −  2³ ; 

a = 3 ;  b = 2 ; 
___________________________________________________________
Plug in these values:

 " a³ − b³ = (a − b)(a² + ab + b²) " ; 

→   3³ − 2³  ≟  [3 − 2) [ 3² + (3*2) + 2² ]  ≟  19 ?  ; 

→  27 − 8   ≟  (1) (9 + 6 + 4) ≟  19 ?  ; 

→     19      ≟     (1) (15 + 4)   ≟  19 ?  ; 

→       19    ≟         (1) (19)     ≟  19 ?  ; 

→         19  ≟                19      ≟    19 ? ; 

→          19  =   19   =  19 !  Yes!  
___________________________________________________________
pickupchik [31]3 years ago
5 0

Answer:

B

Step-by-step explanation:

difference of cubes

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If y=80,when x=32 find x when y=100
Bess [88]
Y        X
80     32
100    x'

80x' = 100*32
80x' = 3200
x' = 3200/80
x' = 320/8
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5 0
3 years ago
What value of z makes the equation z+6÷13=4 true?
netineya [11]

Answer:

z = 46/13

Step-by-step explanation:

Step 1: Write equation

z + 6 ÷ 13 = 4

Step 2: Solve for <em>z</em>

  1. Rewrite: z + 6/13 = 4
  2. Subtract 6/13 on both sides; z = 46/13

Step 3: Check

<em>Plug in z to verify it's a solution.</em>

46/13 + 6/13 = 4

52/13 = 4

4 = 4

6 0
4 years ago
I have a question for special product patterns.<br><br> It is in the attached image.
anastassius [24]
Its answer 3 because they are the only difference with two squared numbers
7 0
3 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
Please show your work for the equation ​
zubka84 [21]

Answer:

y = x + 1

Step-by-step explanation:

We can see the line is linear so we can write the equation using y = mx + b. The components of the equation are m = slope and b = y-intercept. We can see from the graph that the y-intercept is 1 as the graph crosses the y axis at y = 1 and if we do \frac{rise}{run} or \frac{y_{2}  - y_{1} }{x_{2}  - x_{1} }, we can plug in the y-intercept's coorinates and the point in red's coordinates to get the slope of 1. Therefore, the equation would be y = 1x + 1 or y = x + 1

4 0
3 years ago
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