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miskamm [114]
3 years ago
8

How to show the work for x4-y4

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
6 0
I think you can solve it by
=(x^2 - y^2) *( x^2 + y^2)
= (x -y)*(x+y)*(x^2 + y^2)
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PLEASE HELP ASAP!!!!
Paul [167]

Answer:

the answers are <u>B and D </u>

Step-by-step explanation:

please let me know if I am wrong

I found about the answers by determine bias questions from non bias questions.

6 0
4 years ago
DESCRIBE HOW TO WRITE 3 DIGIT NUMBERS IN THREE DIFFERENT WAYS
victus00 [196]
Just put each number in different positions.
425
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3 0
3 years ago
Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati
Brrunno [24]
Remember that <span>an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.
 
Lets solve our equation to find out what is the extraneous solution:
</span>\sqrt{x-3} =x-5
(\sqrt{x-3})^2 =(x-5)^2
x-3=x^2-10x+25
x^2-11x+28=0
(x-4)(x-7)=0
x-4=0 and x-7=0
x=4 and x=7
<span>
So, the solutions of our equation are </span>x=4 and x=7. Lets replace each solution in our original equation to check if they are valid solutions:
- For x=7
\sqrt{x-3} =x-5
\sqrt{7-3} =7-5
\sqrt{4} =2
2=2
We can conclude that 7 is a valid solution of the equation.

- For x=4
\sqrt{x-3} =x-5
\sqrt{4-3} =4-5
\sqrt{1} =1
1 \neq 1
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: <span>D. the extraneous solution is x = 4</span>
7 0
3 years ago
What is the coefficient of xy^2<br> in the expansion of (x+y)^3?
Basile [38]

Step-by-step explanation:

Using Binomial Expansion,

(x + y)³

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Therefore the coefficient of xy² is 3C2 = 3.

5 0
3 years ago
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There are 14 chairs in each
Gnoma [55]

Answer:

37 rows I think has for all

8 0
2 years ago
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