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2 years ago

Which value is equivalent to cos 10°? sin 30° O sin 25° O sin 70° O sin 80°

Mathematics

2 answers:

Sauron [17]2 years ago

8 0

Answer:

Step-by-step explanation:

STatiana [176]2 years ago

7 0

Answer is 13

explanation

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I need help. Thank you.

Oduvanchick [21]

Answer:

A=-5

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago

If the population in a state is 1,821,000 and is decreasing by 0.2% every year. What will the population be in 5 years? Round to

ikadub [295]

1,802,790 is the awnser

3 0

3 years ago

Find the equation of the linear function represented by the table below in slope-

7nadin3 [17]

to get the equation of any straight line we simply need two points off of it, hmm let's get two from this table

$\begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} 0&4\\ 1&10\\ 2&16\\ 3&22\\ 4&28\\ \cline{1-2} \end{array} \begin{array}{llll} \\ \leftarrow \textit{let's use this point}\\\\ \leftarrow \textit{and this point} \end{array}$

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{22}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{22}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{12}{2}\implies 6 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{6}(x-\stackrel{x_1}{1}) \\\\\\ y-10=6x-6\implies y=6x+4$

7 0

1 year ago

Help me again plz quick

madreJ [45]

Answer:

72/18=4

Step-by-step explanation:

3 0

2 years ago