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kiruha [24]
3 years ago
11

a sayian's power level is 400 septillion if he goes super sayian 3 and it multiplies his power by 900 trillion what is his power

level in super sayian 3
Mathematics
2 answers:
QveST [7]3 years ago
7 0
It's over 9 thousand
Akimi4 [234]3 years ago
7 0

Answer:

3.6e+41 the e+ means the amount of 0s i think so


3,600,000,000,000,000,000,000,000,000,000,000,000,000,000


but for the record the super saiyan 3 transformation is supposed to be a 400x multiplier so his actual power level would be 3.6e+29 or 3,600,000,000,000,000,000,000,000,000,000

Step-by-step explanation:

im a db fan and i own a calculator

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Rename 600,000= ten thousands
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For this case we have the following number:

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600,000 = sixty - ten thousand s

Answer:

Rewriting 600,000 we have:

600,000 = sixty - ten thousands

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Answer:

Step-by-step explanation:

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The expression (secx + tanx)2 is the same as _____.
trapecia [35]

<u>Answer:</u>

The expression \bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}

<u>Solution:</u>

From question, given that \bold{(\sec x+\tan x)^{2}}

By using the trigonometric identity (a + b)^{2} = a^{2} + 2ab + b^{2} the above equation becomes,

(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x

We know that \sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}

(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}

=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}

On simplication we get

=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}

By using the trigonometric identity \cos ^{2} x=1-\sin ^{2} x ,the above equation becomes

=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}

By using the trigonometric identity (a+b)^{2}=a^{2}+2ab+b^{2}

we get 1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}

=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}

=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}

By using the trigonometric identity a^{2}-b^{2}=(a+b)(a-b)  we get 1-\sin ^{2} x=(1+\sin x)(1-\sin x)

=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}

= \frac{1+\sin x}{1-\sin x}

Hence the expression \bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}

8 0
3 years ago
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