3 years ago

Find the sum of the smallest 7- digit number and largest 4-digit number

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

7 0

1009999

Hope it helps you...

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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

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

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3 years ago

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Answerr 5 & 6 pleaseee

choli [55]

Answer:

5. 2

6. 10

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Step-by-step explanation:

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Which expression is equivalent to (q^4)^-3/q^-15 for all values of q where the expression is defined?

vodka [1.7K]

Answer: (h)q^3

Step-by-step explanation:

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3 years ago

What are the steps to solve <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B20%20%7Bk%7D%5E%7B3%7D%20%7D%20" id="TexFormula1"

vlada-n [284]

$\bf \sqrt{20k^3}\qquad 
\begin{cases}
20=2\cdot 2\cdot 5\\
\qquad 2^2\cdot 5\\
k^3=k^{2+1}\\
\qquad k^2\cdot k^1
\end{cases}\implies \sqrt{2^2\cdot 5\cdot k^2\cdot k}\implies 2k\sqrt{5k}$