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3 years ago

Which table shows a proportional relationship

Mathematics

1 answer:

Nookie1986 [14]3 years ago

6 0

9514 1404 393

Answer:

see below

Step-by-step explanation:

In a proportional relationship, the ratio of y to x is the same for all values of x and y. Here, you can determine if that is the case by looking at the first two table entries in each table.

A. -4/1 = -12/3 . . . proportional

B. -5/1 ≠ -13/3 . . . not proportional

C. -4/1 ≠ -16/3 . . . not proportional

D. -8/1 ≠ -20/3 . . . not proportional

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Step-by-step explanation:

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2 years ago

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3 years ago

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

MrRissso [65]

$\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3} \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(3x)^{3 \times \frac{5}{3} }}{(2y^3)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}$

Now, If the exponent was negative like you asked....

$\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}$

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2 years ago

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Answer:

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2 years ago