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3 years ago

What Is the ? In the question?

Mathematics

2 answers:

amid [387]3 years ago

6 0

6, it's the only digit from 1-9 that isn't there.

Hope this helps :)

Nikitich [7]3 years ago

4 0

The answer is 6 for the ?

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Write a x (-a)x 13 x a x (-a) x 13 in power notation

miss Akunina [59]

Answer:

$a\times (-a)\times 13\times a\times (-a)\times 13$ can be written in power notation as $a^{4}\times 13^{2}$

Step-by-step explanation:

The given expression

$a\times (-a)\times 13\times a\times (-a)\times 13$

Writing a\times (-a)\times 13\times a\times (-a)\times 13 in power notation:

Let

$a\times (-a)\times 13\times a\times (-a)\times 13$

= $[13\times13][(a\times (-a)\times a\times (-a)]$

$13\times13 = 13^{2}$ , $a\times a = a^{2}$ , $(-a)\times (-a) = (-a)^{2}$

So,

$=[13^{2}][a^2\times (-a)^2]$

$(-a)^2 = a^{2}$

So,

$=[13^{2}][a^2\times a^2]$

As ∵ $a^{m} \times a^{n}=a^{m+n}$

$=[13^{2}][a^{2+2}]$

As ∵ $a^{m} \times a^{n}=a^{m+n}$

$=13^{2}\times a^{4}$

$=a^{4}\times 13^{2}$

Therefore, $a\times (-a)\times 13\times a\times (-a)\times 13$ can be written in power notation as $a^{4}\times 13^{2}$

Keywords: power notation

Learn more about power notation from brainly.com/question/2147364

#learnwithBrainly

5 0

3 years ago

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

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