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horrorfan [7]
2 years ago
8

Find all solutions in the interval [0, 2π). (sin x)(cos x) = 0

Mathematics
2 answers:
Trava [24]2 years ago
5 0

Answer:

The solution in the interval [0, 2π) of the expression (\sin x)(\cos x) = 0 are x=0,\pi,\frac{\pi}{2},\frac{3\pi}{2}                                

Step-by-step explanation:

Given : Expression (\sin x)(\cos x) = 0

To find : All the solutions in the interval [0, 2π) ?

Solution :

Expression (\sin x)(\cos x) = 0

When a\cdot b=0\Rightarrow a=0\text{ or } b=0

\sin x= 0  or \cos x= 0      

x=\sin^{-1}(0)  or x=\cos^{-1}(0)        

The general solution of the equations are

x=0,\pi,2\pi,..  or x=\frac{\pi}{2},\frac{3\pi}{2},..

From 0\leq x

The solutions are          

x=0,\pi  or x=\frac{\pi}{2},\frac{3\pi}{2}    

Therefore, The solution in the interval [0, 2π) of the expression (\sin x)(\cos x) = 0 are x=0,\pi,\frac{\pi}{2},\frac{3\pi}{2}  

Deffense [45]2 years ago
3 0
\bf sin(x)cos(x)=0\implies 
\begin{cases}
sin(x)=0\\
\measuredangle x=sin^{-1}(0)\\
\measuredangle x=0\ ,\ \pi \\
----------\\
cos(x)=0\\
\measuredangle x=cos^{-1}(0)\\
\measuredangle x=\frac{\pi }{2}\ ,\ \frac{3\pi }{2}
\end{cases}
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The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}

Given:

\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25

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t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397

*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

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2 years ago
How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

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\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

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\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2

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Solve for <em>N</em> :

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Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

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