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IgorLugansk [536]

3 years ago

7

Solving Inequalities Independent Practice

1 answer:

Gelneren [198K]3 years ago

3 0

Answer:

X< 9

Step-by-step explanation:

1/3x-6<-3 is correct.

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Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

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

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

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

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago

I need the answer ASAP please

lara [203]

a $\leq$ 7

because 4a + 7 - 7 $\leq$ 35 - 7

equals 4a $\leq$ 28

and 4a/4 $\leq$ 28/4

equals a $\leq$ 7

5 0

3 years ago

Read 2 more answers

What is the area of this triangle

Misha Larkins [42]

Answer:

Probably b or e

Step-by-step explanation:

this took me a while to answer this.

4 0

3 years ago

Which statement about the dilation of these triangles is true?

natulia [17]

Jabardasth question is the scale factor is 2

7 0

3 years ago

Read 2 more answers

Point Q' is the image of Q(-7, -6) under the translation (x, y) + (x + 12, y + 8).

Lina20 [59]

Answer:

The co-ordinates of Q' is (5,2).

Step-by-step explanation:

Given:

Pre-image point

Q(-7,-6)

To find Image point Q' after following translation.

$(x,y)\rightarrow (x+12,y+8)$

Solution:

Translation rules:

Horizontal shift:

$(x,y)\rightarrow (x+k,y)$

when $K>0$ the point is translated $k$ units to the right.

when $K$ the point is translated $k$ units to the left.

Vertical shift:

$(x,y)\rightarrow (x,y+k)$

when $K>0$ the point is translated $k$ units up.

when $K$ the point is translated $k$ units down.

Given translation $(x,y)\rightarrow (x+12,y+8)$ shows the point is shifted 12 units to the right and 8 units up.

The point Q' can be given as:

Q'= $(-7+12,-6+8)=(5,2)$

So, the co-ordinates of Q' is (5,2). (Answer)

7 0

3 years ago