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Yanka [14]
3 years ago
8

If 24, x, and 6 form the first three terms of an arithmetic sequence

Mathematics
1 answer:
AveGali [126]3 years ago
7 0
<h3>Answer:  15</h3>

===============================================

Work Shown:

d = common difference

p = first term = 24

q = second term = a+d = 24+d

r = third term = q+d = 24+d+d = 24+2d = 6

------------

Solve for d

24+2d = 6

2d = 6-24

2d = -18

d = -18/2

d = -9

We add -9 to each term to get the next term. This is the same as subtracting 9 from each term to get the next term.

------------

First term = 24

Second term = 24-9 = 15

Third term = 15-9 = 6

We get the sequence 24, 15, 6

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● For the polygon you are assigned, assume the radius is 1 unit . Round all answers to the nearest hundredth , if necessary.
lorasvet [3.4K]

9514 1404 393

Answer:

  1) 135°

  2) 0.77 units

  3) 0.92 units

  4) 6.12 units

  5) 2.83 square units

Step-by-step explanation:

1) The exterior angle at any vertex is 360°/n, where n is the number of sides. For the octagon, the exterior angle is 360°/8 = 45°. The interior angle will be the supplement of this: 180° -45° = 135°.

The measure of each interior angle is 135°.

__

2) Each sector of the octagon is an isosceles triangle with a central angle of 45° (also 360°/8). Since the radius is 1 unit, the length of one side is twice the sine of half the central angle.

  s = 2·sin(45°/2) ≈ 0.765367

The length of one side is about 0.77 units.

__

3) Similarly, the apothem is the cosine of half the central angle:

  a = cos(45°/2) ≈ 0.923880

The apothem is about 0.92 units.

__

4) For an octagon, the perimeter is 8 times the length of one side.

  P = 8s = 8(0.765367) ≈ 6.12293

The perimeter is about 6.12 units.

__

5) The area of one sector (isosceles triangle) is given by the formula ...

  A = (1/2)bh

  A = 1/2sa ≈ 1/2(0.765367)(0.923880) ≈ 0.353553

Then the area of the octagon is 8 times this:

  A = 8(sector area) = 8(0.353553) ≈ 2.82843

The octagon area is about 2.83 square units.

3 0
3 years ago
Clear<br>The perimeter of a regular octagon is 88 mm. How long is each side?​
Colt1911 [192]

Answer:

11 mm

Step-by-step explanation:

An octagon has 8 sides. This mean that we do 88 divided by 8 to get the length of each side which is 11 mm.

3 0
3 years ago
The width of a rectangle is 5 meters less than its​ length, and the perimeter is 34 meters. Find the length and width of the rec
Alona [7]

Answer:

Length = 11

Width = 6

Step-by-step explanation:

We know that the length is x and the width is x - 5 since it is 5 less than the length.

The equation to find the perimeter of a rectangle is 2w + 2l

We will plug in the values and solve

2(x-5) + 2(x) = 34

2x - 10 + 2x = 34

4x - 10 = 34

4x = 44

x = 11

Since the length is simply x, we know that it is 11. We subtract 5 to find the width.

11 - 5 = 6.

The width is 6.

3 0
3 years ago
What is the mean of 4,6,7,7,11,12,15,18,20
ExtremeBDS [4]

Answer:

<h2>11.1111111111 or 1.1 <-(rounded)</h2>

Step-by-step explanation:

4+6+7+7+11+12+15+18+20= 100

divide by the number of values we have. we have 9 values

100/9=11.1111111111

I hope this helped

4 0
3 years ago
Read 2 more answers
PLEASE HELP!!!
pogonyaev

Step-by-step explanation:

Part A:

Let m be the number of mittens and s be the number of scarves. Then we have the inequalities:

s+m\leq 30. <em>This says Nivyana and Ana cannot make more than 30 scarves</em>

50s+25m\geq 1000. <em>This says that</em> <em>Nivyana and Ana have to earn at least $1000.</em>

Part B:

The graph is attached.

Notice that the graphs of the inequalities are solid lines, this just means that the points on these lines included to the solutions of each inequality.

The darker shaded region and the solid lines bounding it, are the solutions to the inequalities because that's where the values common to both inequalities are found.

Part C:

From the graph we get two possible solutions:

15 scarves & 10 mittens

25 scarves & 5 mittens.

These two points lie on the solid lines that bound the darker shaded region<em> (I picked those points to stress that the lines bounding the dark region are also solutions.)</em>

8 0
3 years ago
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