I need EXTREME HELP PLEASE

Mathematics

1 answer:

elena55 [62]1 year ago

6 0

Answer:

8 months

Step-by-step explanation:

194-3m = 217-6m

3m = 23

m=7.666666..... (8 months

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Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

2 years ago

The quotient of m and 7.

Firdavs [7]

M/7 would be the correct answer

6 0

2 years ago

Let f(x) = tan(x) - 2/x. Let g(x) = x^2 + 8. What is f(x)*g(y)?

Tema [17]

Answer:

$f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}$