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Virty [35]
3 years ago
8

If sin θ = 0.55 what is the value of cos θ? How do I solve this?

Mathematics
1 answer:
givi [52]3 years ago
4 0
0.55^2 + cos^2 = 1
cos^2 = 1 - 0.55^2
cos^2 = 1 - 0.3025
cos^2 = 0.6975
cos = 0.8351
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Answer:

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Step-by-step explanation:

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Answer:

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The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo
oee [108]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

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Step-by-step explanation:

The coordinates forming the polygon are

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Therefore, the length between points AB is

length AB = \sqrt{(-4 - (-2))^2+(-1-3)^2}

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Similarly, length BC is given by

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Length DA = \sqrt{(8)^2+(-2)^2} =  \sqrt{68} = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

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D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A =  \sqrt{(7)^2+(0)^2} =  \sqrt{49} =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

 S_{\bigtriangleup} = (\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|

= (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

S_{{\bigtriangleup}ABC} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

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Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

S_{{\bigtriangleup}ACD'} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

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Area of polygon =   S_{{\bigtriangleup}ABCD'} = S_{{\bigtriangleup}ABC} + S_{{\bigtriangleup}ACD'} = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

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What is the area of a parallelogram?
denis23 [38]
D is the correct answer




the area of a parallelogram is equal to: base x height

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good luck
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Answer:

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