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rosijanka [135]
3 years ago
8

PLEASEE HELP ANYONE WHO IS GOOD AT MATH

Mathematics
2 answers:
Lorico [155]3 years ago
8 0
\sqrt{\frac{126xy^{5}}{32x^{3}}}=\sqrt{\frac{63y^{5}}{ax^{b}}}\\\\\sqrt{\frac{63y^{5}}{16x^{2}}}=\sqrt{\frac{63y^{5}}{ax^{b}}}\\\\\frac{\sqrt{63y^{5}}}{\sqrt{16x^{2}}}=\frac{\sqrt{63y^{5}}}{\sqrt{ax^{b}}}\\\\\frac{\sqrt{63y^{5}}}{\sqrt{16x^{2}}}*\frac{1}{\sqrt{63y^{5}}}=\frac{\sqrt{63y^{5}}}{\sqrt{ax^{b}}}*\frac{1}{\sqrt{63y^{5}}}\\\\\frac{1}{\sqrt{16x^{2}}}=\frac{1}{\sqrt{ax^{b}}}\\\\\sqrt{16x^{2}}=\sqrt{ax^{b}}

Thus for the two equations to be equal, a = 16 and b = 2.
Lapatulllka [165]3 years ago
7 0
What was done from the first to the second equation was that the fraction was simplified. 126 and 32 have a common factors of 2. 
126/32=(63*2)/(16*2)
The 2s in the top and bottom can cancel out, leaving the fraction 63/16. 
In addition, since there are x terms on the top and bottom, they cancelled out as well. 
x/x^3=1/x^2
This leaves an x^2 term on the bottom. 
Thus, if a is 16, and b is 2, you will have an equivalent form of the fraction.
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pav-90 [236]

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Step-by-step explanation:

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Step-by-step explanation:

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In the graph, we see that the equation intersects the x-axis at two distinct points. Therefore, the quadratic has two solutions and the discriminant must be positive. Thus, we have b^2-4ac>0.

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