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Mathematics

1 answer:

olga2289 [7]3 years ago

6 0

X = 4.

This is because on the graph when x is 3 both equations output or y are 3.

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X^2+4x-3x^3+6 in standard form <br><br> show work

Hitman42 [59]

Hello :
the standard form ( X^2+4X-3X^3+6 ) is : -3X^3+X^2+4X+6

4 0

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....

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

5 0

3 years ago

4 < b select all that apply <br> b=8 b=6 b=7 b=10

valentinak56 [21]

Answer:

THE ANSWER IS B=8

Step-by-step explanation:

Beause 4 goes into 8 2 times!!!

6 0

3 years ago

Read 2 more answers

Find anequation for the perpendicular bisector of the line segment whose endpoints are(3, -1) and (-9,5).