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bearhunter [10]
2 years ago
8

Help please you don’t know how much this means to me

Mathematics
1 answer:
ICE Princess25 [194]2 years ago
5 0
<h2>Problem 6:</h2><h2>a)</h2>

a(0) = 1  \:  \:  \:  \:  \:  \: b(0) = 2 \:  \:  \:  \:  \: c(0) = 3 \\ a(1) = b(0) + c(0) = 2 + 3 = 5 \\ b(1) = a(0) + c(0) = 1 + 3 = 4 \\ c(1) = a(0) + b(0) = 1 + 2 = 3 \\  \\ a(2) = b(1) + c(1) = 4 + 3 = 7 \\ b(2) = a(1) + c(1) = 5 + 3 = 8 \\ c(2) = a(1) + b(1) = 5 + 4 = 9

a(3) = b(2) + c(2) =  8 + 9 = 17\\ b(3) = a(2) + c(2) =7 + 9 = 16  \\ c(3) = a(2) + b(2) = 7 + 8 = 15 \\  \\ a(4) = b(3) + c(3) = 16  + 15 = 31 \\ b(4) = a(3) + c(3) = 17 + 15 = 32 \\ c(4) = a(3) + b(3) = 17 + 16 = 33

a(5) = 32 + 33 = 65 \\ b(5) = 31 + 33 = 64 \\ c(5) =31 + 32 =  63 \\  \\ a(6) = 64 + 63 = 127 \\ b(6) = 65 + 63 = 128 \\ c(6) = 65 + 64 = 129 \\

a(7) = 128 + 129 = 257 \\ b(7) = 127 + 129 = 256 \\c (7) = 127 + 128 = 255 \\  \\ a(8) = 256 + 255 = 511 \\ b(8) = 257 + 255 = 512 \\ c(8) = 257 + 256 = 513

a(9) = 512 + 513 = 1025 \\ b(9) = 511 + 513 = 1024 \\ c(9) = 511 + 512 = 1023 \\  \\ a(10) = 1024 + 1023 = 2047 \\ b(10) = 1025  + 1023 = 2048 \\ c(10) = 1025 + 1024 = 2049

<h2>b)</h2>

a(n) + b(n) + c(n) =   \\  2(a(n - 1) + b(n - 1) + c(n - 1)) \\   6 \times 2 {}^{n }

<h2>c)</h2>

6 \times 2 {}^{n}  > 100 \: 000 \\ 2 {}^{n}  >  \frac{100 \: 000}{6}  \\ n >  log {}^{2} ( \frac{100 \: 000}{6} )  \\ n > 14.02468 \\ n = 15

<h2 />
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Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

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