2 years ago

Please help me with these factoring problems. Please help me solve. I can't do these!!

1 answer:

3 0

Simple...

you have:

1.) -8x-16 --->>>

Factor out a -8 -->>

-8(x+2)

2.)9d+15db --->

Factor out 3d -->

3d(3+5b)

3.) 2a-9ax-->>

Factor out -a -->

-1a(9x-2)

Thus, your answer.

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Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Oduvanchick [21]

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$ , the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$ , so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$