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3 years ago

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1 answer:

Burka [1]3 years ago

6 0

Answer:

19,27 cm^2

Step-by-step explanation:

area of triangle = (6 x 3)/ 2 = 9 cm^2

circle area : 3^2 x 3,14 = 28,27 cm^2

area of shaped region = 28,27 - 9 = 19,27 cm^2

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How to find a tangent ratio

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Answer:

The tangent ratio is the value received when the length of the side opposite of angle theta is divided by the length of the side adjacent to angle theta. It is very commonly abbreviated as tan.

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3 years ago

A rectangle and a triangle have the same area. If their bases are the same length's, how do their lives compare? Justify your an

kozerog [31]

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3 years ago

I) Express Z1 in polar Form<br> a) Z1=1-(2-√3)i (complex number)

kicyunya [14]

Step-by-step explanation:

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7 0

2 years ago

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

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,

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

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.

6 0

3 years ago

I need help ASAP please

galina1969 [7]

I honestly don’t know

Sorry sweety

6 0

3 years ago

Read 2 more answers