$(-3c^{-3}w^5)^3\qquad\text{use}\ (ab)^n=a^nb^n\\\\=(-3)^3(c^{-3})^3(w^5)^3\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=-27c^{-3\cdot3}w^{5\cdot3}=-27c^{-9}w^{15}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=-27\left(\dfrac{1}{c^9}\right)w^{15}=\dfrac{-27w^{15}}{c^9}$

4 0

3 years ago

The first three terms of a sequence are 6, 12, 30, . . .

marta [7]

Answer:

306

Step-by-step explanation:

Take a list of primes and multiply each with the prime+1:

2*3 = 6

3*4 = 12

5*6 = 30

7*8 = 56

11*12 = 132

13*14 = 182

17*18 = 306

8 0

2 years ago

SHOW WORK solve for x 12x>9(2x-3)-15

Thepotemich [5.8K]

12x > 18x - 27 - 15, 12x > 18x - 42, 0 > 6x - 42, 42 > 6x, 7 > x, x < 7

5 0

3 years ago

Read 2 more answers

Solve : x^2+(x-8)^2-3(x+2)=x(x+9)

Tems11 [23]

Answer: x=14+√138,14−√138

6 0

2 years ago