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Ierofanga [76]
3 years ago
8

Help please!! Solve for x, show all work. Thanks!

Mathematics
1 answer:
ELEN [110]3 years ago
8 0

Answer: 54

Step-by-step explanation:

The second term 3/2 loga 9 = loga 9^3/2 = loga 27

The RHS = loga 27 x 2 = loga 54

So loga x = loga 54

x = 54

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katen-ka-za [31]

Answer: <em>(1,5,7,10,11,15)</em>

To get the union of two sets of numbers, all you do is combine the numbers from least to greatest. If a number repeats in the 2 sets, you only put it once.

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List three exampls of a rate
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Step-by-step explanation:

It is a special type of ratio

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Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b
trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) \cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}

( 2 ) \sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}

These two identities makes sin(π / 10) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and cos(π / 10) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}.

Therefore cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}. Substitute,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right]

And now simplify this expression to receive our answer,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right] = -\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i,

-\frac{3\sqrt{5+\sqrt{5}}}{4} = -2.01749\dots and \:\frac{3\sqrt{3-\sqrt{5}}}{4} = 0.65552\dots

= -2.01749+0.65552i

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}

We know that 6\sqrt{5+\sqrt{5}} = 16.13996\dots and -\:6\sqrt{3-\sqrt{5}} = -5.24419\dots . Therefore,

Solution : 16.13996 - 5.24419i

Which rounds to about option b.

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