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3 years ago

Mia found the area of a polygon The area is 32 square cm.

Mathematics

1 answer:

solniwko [45]3 years ago

8 0

Answer:

Only the isosceles trapezoid has an area of 32 cm².

Step-by-step explanation:

Let's calculate the area of each polygon.

For the two triangles we have:

$A_{t} = \frac{bh}{2} = \frac{2 cm*4 cm}{2} = 4 cm^{2}$

This polygon does not have an area of 32 cm².

For the rectangle we have:

$A_{r} = bh = 6 cm*4 cm = 24 cm^{2}$

This polygon does not have an area of 32 cm².

For the rectangle trapezoid we have:

$A_{rt} = A_{t} + A_{r} = (4 + 24) cm^{2} = 28 cm^{2}$

So, this polygon does not have an area of 32 cm².

Finally, for the isosceles trapezoid:

$A_{it} = A_{t}*2 + A_{r} = (4*2 + 24) cm^{2} = 32 cm^{2}$

This polygon does have an area of 32 cm².

Therefore, only the isosceles trapezoid has an area of 32 cm².

I hope it helps you!

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3 years ago

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

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<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$