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3 years ago

14

Write a whole number and a fraction greater than i to name the part filled. Think: 1 container

2 answers:

Anit [1.1K]3 years ago

8 0

8/4 or 2............
hope this helped:)

rusak2 [61]3 years ago

4 0

Answer:

Step-by-step explanation:

There are two containers in which four parts has to be filled in each, therefore, the fraction greater than one will be= $\frac{Total number of parts to be filled in containers}{Number of parts in each container}$

= $\frac{8}{4}$

The whole number greater than one will be= 2

Hence, the whole number and a fraction greater than one to name the part filled is 2 and $\frac{8}{4}$ .

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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,

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

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

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

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

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

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AveGali [126]

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