Get graph paper make a graph mark them on your graph and since it says (n,-2) its going down because of the negative 2 i don't think this helped but if it did

7 0

3 years ago

one moving company charges $800 plus $16 per hour another moving company charges $720 plus $21 an hour at what number of hours w

Gekata [30.6K]

800 + 16x = 720 + 21x. Cancel out the 720 to get 80 + 16x = 21x. Subtract 16x from both sides to get 80 = 5x. Divide by 5 to get 16 = x, or x = 16. 16 hours. The charge is 800 + 16(16), = 720 + 21(16) which = 800 + 256 = 720 + 336 which = 1,056 = 1,056. Hope this helped!

6 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

Timothy brought in brownies for his class on his birthday.

IgorC [24]

Answer:

Incomplete problem - reword

Step-by-step explanation:

8 0

2 years ago