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3 years ago

This composite figure is made up of three simpler shapes. What is the area of

Mathematics

1 answer:

Lady_Fox [76]3 years ago

4 0

Answer:

B. 175 square cm

Step-by-step explanation:

Area of the figure = Area of parallelogram + Area of square + Area of triangle

Area of parallelogram = (base x height)

= (14 x 5)

= 70 $cm^{2}$

Area of square = (length x length)

= (7 x 7)

= 49 $cm^{2}$

Area of triangle = ( $\frac{1}{2}$ base x height)

= ( $\frac{1}{2}$ x 16 x 7)

= 56 $cm^{2}$

Area of the figure = 70+ 49 + 56

= 175

The area of the figure is 175 $cm^{2}$ .

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raketka [301]

Answer:

B. People who were promised free samples of Penelope‘s puffy popcorn for completing the survey

Step-by-step explanation:

This would show likely bias in the result of the survey

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2 years ago

How many times greater is the volume of the sphere than the volume of cone #1? Round your answer to the neared tenth. Use 3.14 f

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3 years ago

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4 years ago

A motorboat travels 165 kilometers in 3 hours going upstream and 510 kilometers in 6 hours going downstream. What is the rate of

nikitadnepr [17]

<h3>Rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr</h3><h3><u>Solution:</u></h3>

Given that,

A motorboat travels 165 kilometers in 3 hours going upstream and 510 kilometers in 6 hours going downstream

Therefore,

Upstream distance = 165 km

Upstream time = 3 hours

<h3><u>Find upstream speed:</u></h3>

$speed = \frac{distance}{time}\\\\speed = \frac{165}{3}\\\\speed = 55$

Thus upstream speed is 55 km per hour

Downstream distance = 510 km

Downstream time = 6 hours

<h3><u>Find downstream speed:</u></h3>

$speed = \frac{510}{6}\\\\speed = 85$

Thus downstream speed is 85 km per hour

<em><u>If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then</u></em>

Speed downstream = u + v km/hr

Speed upstream = u - v km/hr

Therefore,

u + v = 85 ----- eqn 1

u - v = 55 ----- eqn 2

Solve both

Add them

u + v + u - v = 85 + 55

2u = 140

u = 70

<em><u>Substitute u = 70 in eqn 1</u></em>

70 + v = 85

v = 85 - 70

v = 15

Thus rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr

3 0

3 years ago

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