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3 years ago

23^0 is equal to O 23 1 23

Mathematics

2 answers:

maksim [4K]3 years ago

7 0

Answer:

23^0=0

Step-by-step explanation:

Anything to the power of 0 equals 0 because it is that number 0 times.

Hope this helps! Plz give brainliest!

juin [17]3 years ago

5 0

Answer:

23^0 is equal to

The Correct Answer is 1

Hope this helps!

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Answer:

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Step-by-step explanation:

To find your answer, simply plug in Francesca's time and Phil's time to your cost equation.

Francesca can be represented by the following equation:

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Phil can be represented by the following equation:

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As such, Francesca pays $13 and Phil pays $39. Next, since you want to find how much more Phil paid, all you have to do is subtract 13 from 39. That will give you $26.

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Answer:

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Step-by-step explanation:

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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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Answer:

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