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cestrela7 [59]
2 years ago
13

(2m+7)^3 written in standard form

Mathematics
1 answer:
elena-s [515]2 years ago
6 0

Answer:

2

m

2

+

7

m

−

3

Step-by-step explanation:

2

m

2

+

7

m

−

3

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andriy [413]

Area of the figure = 30.28 m²

Solution:

The given image is splitted into two shapes.

One is rectangle and the other is semi-circle.

Length of the rectangle = 6 m

Width of the rectangle = 4 m

Area of the rectangle = length × width

                                    = 6 m × 4 m

                                    = 24 m²

Area of the rectangle = 24 m²

Diameter of the semi-circle = 4 m

Radius of the semi-circle = 4 m ÷ 2 = 2 m

Area of the semi-circle = \frac{1}{2} \ \pi r^2

                                      $=\frac{1}{2} \times 3 .14 \times 2^2

                                      =6.28

Area of the semi-circle = 6.28 m²

Area of the figure = Area of the rectangle + Area of the semi-circle

                              = 24 m² + 6.28 m²

                              = 30.28 m²

Area of the figure = 30.28 m²

3 0
3 years ago
Which of these problem types can not be solved using the Law of Sines?
andrezito [222]

I think the answer is B!

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AlladinOne [14]

<em>*To solve for a specified variable, you need to isolate that variable onto one side.</em>

<h3>11.</h3>

Firstly, subtract 5r on both sides: 2p=q-5r

Lastly, divide both sides by 2 and <u>your answer will be p=\frac{1}{2}q-\frac{5}{2}r</u>

<h3>12.</h3>

First, subtract z on both sides of the equation: -10-z=xy

Next, divide both sides by y and <u>your answer will be \frac{-10-z}{y}=x</u>

<h3>13.</h3>

Firstly, multiply both sides by b: a=cb

Next, divide both sides by c and <u>your answer will be \frac{a}{c}=b</u>

5 0
3 years ago
Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha
GalinKa [24]

Answer:

\theta_1 \ n\ \theta_2 = 120, 240

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

Nonagon = 9\ sides

Hexagon = 6\ sides

Divide 360 degrees by the number of sides in each angle, then find the multiples.

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\theta = \frac{360}{9} =40

List the multiples of 40

\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320

<u>Hexagon</u>

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List the multiples of 60

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List out the common angles

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\theta_2 = 60, 120, 180, 240, 300

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The other angles will only work on one of the shapes, but not both at the same time.

7 0
3 years ago
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