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3 years ago

What is the solution to the system below? 3x + y = 11 y = x + 3

Mathematics

1 answer:

alexgriva [62]3 years ago

5 0

Answer:

x = 2

y = 5

Step-by-step explanation:

3x + y = 11 equation (1)

-x + y = 3 multiply this equation by -1

x - y = -3 equation (2)

by adding 2 equations

4x = 8

x = 2

by sub. in Equation (2)

2 - y = -3

y = 2+3

y = 5

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Sholpan [36]

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,

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

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

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

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

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Therefore, the correct option is B.

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