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3 years ago

5

I answered all my work correctly but I don’t understand this one.

1 answer:

SpyIntel [72]3 years ago

3 0

Take your x values in each coordinate and subtract 2, and take your y values and subtract 1.

Q (0-2), (-1-1)
= (-2,-2)

D (-2-2), (2-1)
= (-4, 1)

V (2-2), (4-1)
= (0,3)

J (3-2), (0-1)
= (1,-1)

You can also draw it on a graph and then translate all coordinates 2 units left and 1 down to see the end results.

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Convince me why points B&C are solutions to this system of linear inequalities.

muminat

The coordinates of B and C are solutions to the system of inequalities because both are true to the two inequalities. A is not a solution to the system of inequalities because it is only true to “y is less than or equal to 2x”.

3 0

3 years ago

Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin<0

Gre4nikov [31]

Answer:

Using trigonometric ratio:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

From the given statement:

$\cos \theta = -\frac{4}{5}$ and sin < 0

⇒ $\theta$ lies in the 3rd quadrant.

then;

$\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

Using trigonometry identities:

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

Substitute the given values we have;

$\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}$

Since, sin < 0

⇒ $\sin \theta = -\frac{3}{5}$

now, find $\tan \theta$ :

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

Substitute the given values we have;

$\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}$

Therefore, the exact value of:

(a)

$\sec \theta =-\frac{5}{4}$

(b)

$\tan \theta= \frac{3}{4}$

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