All categories

3 years ago

12(-5z - 4) please help me

Mathematics

1 answer:

damaskus [11]3 years ago

7 0

Answer:

-60z - 48

Step-by-step explanation:

Is your expression.

You might be interested in

In any circle, the ratio of the _ to the radius is 2.

erica [24]

Answer:

The answer should be C

Step-by-step explanation:

5 0

3 years ago

What is 3 to the 8th power times 3 to the 2nd power simplified and what law did you use?

hodyreva [135]

Answer:

$3^{10}$

Step-by-step explanation:

$3^{8}*3^{2}=3^{8+2}=3^{10}$

Using the multiplication/product law of exponents, you can add the exponents up.

4 0

4 years ago

Read 2 more answers

Find the cube roots of 27(cos 330° + i sin 330°)

Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

DeMoivre's Theorem

Let $z=r(cos\theta+isin\theta)$ be a complex number in polar form, where $n$ is an integer and $n\geq1$ . If $z^n=r^n(cos\theta+isin\theta)^n$ , then $z^n=r^n(cos(n\theta)+isin(n\theta))$ .

Nth Root of a Complex Number

If $n$ is any positive integer, the nth roots of $z=rcis\theta$ are given by $\sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}}$ where the nth roots are found with the formulas:

$\sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr]$ for degrees (the one applicable to this problem)
$\sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr]$ for radians

for $k=0,1,2,...\:,n-1$

Calculation

$z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)$

First cube root where k=2

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]$

Second cube root where k=1

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]$

Third cube root where k=0

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]$