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Xelga [282]
3 years ago
15

For this graph, mark the statements that are true.

Mathematics
1 answer:
Arada [10]3 years ago
7 0
The only one that’s true there is A. Domain refers to x values and range is y values. Your domain is between -1 and 3 and your range starts at zero and increases infinitely upward.
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A/ab-bsquare + b/ab-asquare?​
lord [1]

Answer:

\dfrac{(a+b)}{ab}

Step-by-step explanation:

The given expression is :

\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}

It can be solved as follows :

\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}=\dfrac{a}{b(a-b)}+\dfrac{b}{a(b-a)}\\\\=\dfrac{a}{b(a-b)}+\dfrac{b}{-a(-b+a)}\\\\=\dfrac{1}{a-b}(\dfrac{a}{b}-\dfrac{b}{a})\\\\=\dfrac{a^2-b^2}{ab(a-b)}\\\\=\dfrac{(a-b)(a+b)}{ab(a-b)}\\\\=\dfrac{(a+b)}{ab}

So, the solution of the given expression is equal to \dfrac{(a+b)}{ab}.

7 0
3 years ago
Please help me out please
AlexFokin [52]
Does not exist. The line would be vertical

7 0
3 years ago
Read 2 more answers
Is the following a linear equation ?
NNADVOKAT [17]
I think the answer is B but I am not sure
7 0
3 years ago
HELP PLEASE! I'm not sure what the answer is.
Anuta_ua [19.1K]
The answer for this are (10,2), (8,1), and (14,-1).
7 0
3 years ago
Read 2 more answers
Help me fjfjsncjdhdhfh​
Deffense [45]

Step-by-step explanation:

=>  2 {x}^{2}  +  \frac{7}{2}x +  \frac{3}{4}  = 0

=> \frac{8 {x}^{2}  + 14x + 3}{4} = 0

=> 8 {x}^{2}  + 14x + 3 = 0

=> 8 {x}^{2}  + 12x + 2x + 3 = 0

=> 4x(2x + 3) + 1(2x + 3) = 0

= > (2x + 3)(4x + 1) = 0

=  > x =  -  \frac{3}{2} or \frac{ - 1}{4}

option C

8 0
3 years ago
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