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2 years ago

6.04x10^-3 as an ordinary number

Mathematics

1 answer:

Sergio039 [100]2 years ago

4 0

Answer:

0.00604

Step-by-step explanation:

The exponent tells us how many spaces to move the decimal in a given number. If its positive, the decimal moves to the right. If its negative, the decimal moves to the left.

6.04 x 10⁻³ = 0.00604

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Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

we can simplify left side and make it equal to right side

we can use trig identity

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

$\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}$

$\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}$

$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

$\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can use trig identity

$sin^2(a)+cos^2(a)=1$

$\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}$

we can cancel terms

$=3-4(1-sin(a)cos(a))$

now, we can simplify it further

$=3-4+4sin(a)cos(a))$

$=-1+4sin(a)cos(a))$

$=4sin(a)cos(a)-1$

$=2\times 2sin(a)cos(a)-1$

now, we can use trig identity

$2sin(a)cos(a)=sin(2a)$

we can replace it

$=2sin(2a)-1$

so,

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

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Goshia [24]

1/1
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tensa zangetsu [6.8K]

0.63 is the correct answer

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Mariulka [41]

Answer:

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Answer:

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