16. 5x^3 y^-5 • 4xy^3
20x^4y^-2
20x^4 • 1/y^2
=20x^4/y^2
17. -2b^3c • 4b^2c^2
= -8b^5c^3
18. a^3n^7 / an^4 (a^3 minus a = a^2 same as n^7 minus n^4 = n^3)
=a^2n^3
19. -yz^5 / y^2z^3
= -z^2/y
20. -7x^5y^5z^4 / 21x^7y^5z^2 (divide -7 to 21 and minus xyz)
= -z / 3x^2
21. 9a^7b^5x^5 / 18a^5b^9c^3
=a^2c^2 / 2b^4
22. (n^5)^4
n ^5 x 4
=n^20
23. (z^3)^6
z ^3 x 6
=z^18
Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
Answer:
5<4
Step-by-step explanation:
This inequality is not correct though.
5-7+7<-3+7
5<4
Answer:
No
Step-by-step explanation:
the absolute value function always returns a positive value. The value inside the bars can be positive or negative but the result is always positive
| - 14 | = 14 > - 27
Step-by-step explanation:
4(3x-2) + 6x(2-1)
10x + 11x
21x
