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Neporo4naja [7]
3 years ago
5

The digit with the greatest value 6789

Mathematics
2 answers:
alex41 [277]3 years ago
8 0
The digit with the greatest value is 6 because it is in the thousands place.

6*1000=6000
7*100=700
8*10=80
9*1=9

Of these values, 6000 is the largest due to its place value.
Anarel [89]3 years ago
7 0

Answer:

The digit with the greatest value is 6 because it is in the thousands place.

6*1000=6000

7*100=700

8*10=80

9*1=9

Of these values, 6000 is the largest due to its place value.

Step-by-step explanation:

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8/x-5 - 9/x-4 = 5/x^2-9x+20
adelina 88 [10]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2752942

_______________


Solve the equation:

\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}


Reduce the fractions at the left side so that they have the same denominator:

\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\&#10;\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\&#10;\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\&#10;\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}


Numerators must be equal:

\mathsf{8x-32-(9x-45)=5}\\\\&#10;\mathsf{8x-32-9x+45=5}\\\\&#10;\mathsf{8x-9x=5+32-45}\\\\&#10;\mathsf{-x=-8}\\\\&#10;\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}


I hope this helps. =)


Tags:  <em>rational equation fraction solution algebra</em>

7 0
3 years ago
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