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ratelena [41]
3 years ago
6

PLZ HURRY IT'S URGENT!!!!!

Mathematics
1 answer:
inn [45]3 years ago
6 0

Answer:

the answer is b

Step-by-step explanation:

x + 2/3 = -4

x+ (3) 2/3 = (3) -4

x + 2= -12

x= -14

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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}

5 0
3 years ago
How much is 0.97 to 0.465?
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4 years ago
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