All categories

2 years ago

Someone please help me with this lol… have no idea what I’m doing

Mathematics

1 answer:

Sholpan [36]2 years ago

6 0

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

$\sin \theta=-\sqrt{\dfrac{16}{25}}$

$\sin \theta=-\dfrac{4}{5}$

Therefore, the correct option is B.

You might be interested in

If b is the midpoint of line AC, and AC = 8x-20, find BC.

Elodia [21]

The Definition Of a Midpoint says that is B is the

midpoint of segment AC, then segment AB ≅ AC.

Since segment AB ≅ segment BC, we can setup

the equation 2(3x - 1) = 8x - 20 and we can solve from here.

Start by distributing the 3 through the parenthses on the left side.

So we get 6x - 2 = 8x - 20.

Now subtract 8x from both sides to get -2x - 2 = -20.

Now add 2 to both sides and we have -2x = -18.

Finally, dividing both sides by -2, we find that x = 9.

If x = 9, segment BC must be 3(9) - 1 or 27 - 1 which is 26.

6 0

3 years ago

Angela is buying towels for her apartment. She finds some

sashaice [31]

She can buy 627247463736 towels because she’s rich

3 0

3 years ago

Read 2 more answers

HELPPPPOMOGMGOGMGOGKM PLS ANSWER THIS CORRECTLY PLZZZ THIS IS FOR A VERY IMPORTANT TEST YOUWILL GET BRAINLIEST HELP

allochka39001 [22]

Answer:

i cant give u the answer but here is an example hope this helps slope = "rise over run" the difference in y values divided by the difference in x values

= 6/-3 = -2 = m. It's negative because as x increases, y decreases

plug that into the point slope formula, with either point. generally use the most simple point

y-h= m(x-k) where m= -2 and (k,h) = (2,-2)

y+2 = -2(x-2)

simply if you want

Y = -2x +4 -2 = -2x +2

2x +y = 2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Y=0.5x find the slope and the y-intercept

Vladimir79 [104]

Y=mx+b
m= slope so slope= .5
b= y intercept so y intercept=0

3 0

3 years ago

Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

y''(t) + p(t) y'(t) + q(t) y(t) = f(t)

so the first thing you need to do is divide both sides of your equation by t :

y'' + (2t - 1)/t y' - 2/t y = 7t

You're looking for a solution of the form

$y=y_1u_1+y_2u_2$

where

$u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

$u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

and W denotes the Wronskian determinant.

Compute the Wronskian:

$W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}$

Then

$u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t$

$u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}$

The general solution to the ODE is

$y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}$

which simplifies somewhat to

$\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}$

4 0

3 years ago