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Vedmedyk [2.9K]
4 years ago
12

What is 4.50 in standard form

Mathematics
2 answers:
Valentin [98]4 years ago
7 0
450×10^ -2 = 4.5 is the answer
vlada-n [284]4 years ago
3 0

4.50 in standard form is 450*10^-2

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What's 45,122 round to the nearest ten
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45,120 if its ten.
45,000 if its ten thousand
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Sean X,Y y Z subconjuntos de un conjunto universo U¿El siguiente enunciado es válido?
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The translation is Let X,Y and Z be subsets of a universe set U, Are the following statement valid?

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2 years ago
Lowering powers write in terms of first power of cosine. Cos^6
yaroslaw [1]

The main identity you need is the double angle one for cosine:

\cos^2x=\dfrac{1+\cos2x}2

We get

\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8

Expand the numerator to apply the identity again:

\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8

\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8

\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x

Finally, make use of the product identity for cosine:

\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2

so that ultimately,

\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x

\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x

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