The principal argument of z = —4+i4√3 is​

Let z=−4+0i. Then, ∣z∣=

(−4)

2

+0

=4

Clearly, the point (−4,0) representing z=−4+0i lies on the negative side of real axis. Therefore, principal argument of z is π.

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

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