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Drupady [299]
3 years ago
8

Duran is making party favors by placing a circular disk in the bottom of a plastic bag. Each disk has diameter of 6 centimeters.

What is the area of the disk to nearest hundredth. ** Use 3.14
A. 18.84

B. 28.26

C. 91.06

D. 113.04
Mathematics
1 answer:
Setler79 [48]3 years ago
8 0

Answer:

B. 28.26

Step-by-step explanation:

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Convert 0.0855 to a fraction. Move numbers to the blanks to complete the fraction.
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Answer:

a

Step-by-step explanation:

no

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Gelneren [198K]
13+13=26
9+9=18
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Maria rotated the triangle 90 degrees clockwise about the origin. What is the new triangle?
padilas [110]

Answer:

A’B’C’

Step-by-step explanation:

Well if triangle ABC is rotated 90 degrees clockwise around the origin it will turn right to create triangle A’B’C’.

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3 years ago
Graph the system of equations. {x−y=64x+y=4
Fantom [35]
X-y=6 Equation 1
x+y=4 Equation 2

To graph the given system of equation, first find x and y-intercept of each equation.
x-y=6
When y=0
x=6  Point is (6,0)
When x=0
-y=6
y=-6  Point is (0,-6)

Now x-intercept and y-intercept for equation 2.
x+y=4
When x=0
y=4  Point is (0,4)
When y=0
x=4 Point is (4,0)

Now plot these points on the graph, the lines intersect each other at point (5,-1), which is the solution of the given system.

Answer: (5,-1)

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3 years ago
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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