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natima [27]
2 years ago
7

How are special right triangles used in finding the area of a regular hexagon?

Mathematics
1 answer:
musickatia [10]2 years ago
6 0

Answer:

To find the area of the hexagon,we will first find the area of each equilateral triangle then multiply it by 6.

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Identify the coefficient in the term 7 x2y3 .
Ivanshal [37]

7 x^2y^3

The coefficient is 7

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3 years ago
1. Kaleb worked at a toy store during his summer vacation from
aleksandr82 [10.1K]

Answer:

1. $10.56

Step-by-step explanation:

950.40÷90=10.56

7 0
3 years ago
27 as a mixed number
Elodia [21]

Answer:

you can't do that since its a whole number

4 0
4 years ago
How do I solve this?
andreev551 [17]

Answer:

x is in the range [-1,4]

Step-by-step explanation:

I haven't worked with absolute value inequalities in awhile, but let's take a wack at this.

We are given the following inequality:

| 2x - 3 | <= 5

This implies two possible cases:

[1] -5 <= 2x -3

Or

[2] 2x - 3 <= 5

So let's solve x for both of these cases:

[1] -5 <= 2x - 3

 -2 <= 2x

 -1  <= x

[2]  2x - 3 <= 5

 2x <= 8

  x <= 4

So from these cases, we can say the following is true:

x >= -1  and x <= 4

Thus, we can write this in the form

-1 <= x <= 4

Or in interval notation:

{ x is element of reals | -1 <= x <= 4}

Also written as

x is in the range [-1,4]

Where the closed brackets represent 1 and 4 as possible answers whereas parenthesis would imply they were not.

Cheers.

4 0
3 years ago
M∠Q = <br> m∠R = <br> m∠S = <br><br> 40 POINTS!!!!!!!
Marizza181 [45]

9514 1404 393

Answer:

  ∠Q = 89°

  ∠R = 123°

  ∠S = 91°

Step-by-step explanation:

It seems easiest to start by finding the measures of each of the arcs. The measure of an arc is double the measure of the inscribed angle it subtends.

  arc QRS = 2·∠P = 114°

So, ...

  arc QR = arc QRS - arc RS = 114° -41° = 73°

The total of the arcs around the circle is 360°, so ...

  arc PQ = 360° -arc PS -arc QRS

  arc PQ = 360° -137° -114° = 109°

__

  ∠Q = (1/2)(arc RS + arc PS) = (1/2)(41° +137°)

  ∠Q = 89°

__

  ∠R = (1/2)(arc PS +arc PQ) = (1/2)(137° +109°)

  ∠R = 123°

__

  ∠S = (1/2)(arc PQ +arc QR) = (1/2)(109° +73°)

  ∠S = 91°

7 0
3 years ago
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