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Mkey [24]
3 years ago
13

I need help with my rsm hwrk pls help me

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
3 0

Answer: A= 32(\pi-2)\  cm^2, \ \ \ P=8\pi\ cm

Step-by-step explanation:

When we consider Area under arc AC is is representing a quarter as ADBC is a square, \angle D=90^{\circ}.

Area of quadrant = \dfrac{\pi r^2}{4}

here r= 8 cm

Area under Arc AC= \dfrac{\pi (8)^2}{4}=16\pi\ cm^2

Area of white region ABC = Area of square ADBC - Area under Arc AC

=8^2-16\pi\ \     [\text{Area of square} = sides^2]\\\\= 64-16\pi\ \  =16(4-\pi)\ cm^2

Similarly , Area of white region ADC =  16(4-\pi)\ cm^2

Area of shaded region = Area of square - Area of white region ABC - Area of white region ADC

=64-(64-16\pi)-(64-16\pi)\\\\= 32\pi-64\ cm^2 \\\\= 32(\pi-2)\  cm^2

Area of shaded region == 32(\pi-2)\  cm^2

Length of arc AC = \dfrac{Circumference}{4}=\dfrac{2\pi r}{4}

\dfrac{\pi (8)}{2}=4\pi cm

Perimeter of shaded region = 2(AC) = 8\pi\ cm

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Travis is at a carnival and has a chance to win a goldfish. He can either roll 2 or 3 on a six-sided fair die or pick either an
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Answer:

Travis should pick to play the hat game.

Step-by-step explanation:

You would need to evaluate the probablity of winning in each option

Option 1 (rolling die)

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p(pick A or B) = 2/5

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Since option 2 with picking letters has a higher chance of winning, then Travis should play the second game to win the goldfish as a higher chance of winning makes it easier for him to win the prize.

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Read 2 more answers
Use the unit circle to evaluate these expressions:
sergeinik [125]

Answer:

a) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to 2\pi:

17 \pi/4 - 2\pi = \frac{9\pi}{4} -2\pi = \pi/4

For this case we know that sin (\pi/4) = \frac{\sqrt{2}}{2}

So then sin(\frac{17 \pi}{4}) = \frac{\sqrt{2}}{2}

b) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to 2\pi:

19 \pi/6 - 2\pi = \frac{7\pi}{6}

For this case we know that cos (\pi/6) = \frac{\sqrt{3}}{2}

And we know that \frac{7\pi}{6} is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then cos(\frac{19 \pi}{6}) = -\frac{\sqrt{3}}{2}

c) For this case that any factor of \pi the sin function is equal to 0.

So from definition of tan we have this:

tan (450\pi) = \frac{sin(450 \pi)}{cos(450 \pi)}= \frac{0}{cos(450\pi)}= 0

Step-by-step explanation:

a. sin (17pi / 4 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to 2\pi:

17 \pi/4 - 2\pi = \frac{9\pi}{4} -2\pi = \pi/4

For this case we know that sin (\pi/4) = \frac{\sqrt{2}}{2}

So then sin(\frac{17 \pi}{4}) = \frac{\sqrt{2}}{2}

b. cos (19pi / 6 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to 2\pi:

19 \pi/6 - 2\pi = \frac{7\pi}{6}

For this case we know that cos (\pi/6) = \frac{\sqrt{3}}{2}

And we know that \frac{7\pi}{6} is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then cos(\frac{19 \pi}{6}) = -\frac{\sqrt{3}}{2}

c. tan(450pi)

For this case that any factor of \pi the sin function is equal to 0.

So from definition of tan we have this:

tan (450\pi) = \frac{sin(450 \pi)}{cos(450 \pi)}= \frac{0}{cos(450\pi)}= 0

4 0
3 years ago
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