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lubasha [3.4K]
3 years ago
13

What is the area of the regular hexagon? This is due tomorrow please help! :)

Mathematics
1 answer:
tester [92]3 years ago
3 0

Check the picture below.

so then, the perimeter of that hexagon will  just be the sum of all its 6 sides, or namely 3⅖ + 3⅖ + 3⅖ + 3⅖ + 3⅖ + 3⅖, or just 6( 3⅖ ).

\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=3\\ p=6\left(3\frac{2}{5} \right) \end{cases}\implies A=\cfrac{1}{2}(3)\left[ 6\left(3\frac{2}{5} \right) \right]\implies A=\cfrac{1}{2}(3)\left[ 6\left(\cfrac{17}{5} \right) \right] \\\\\\ A=\cfrac{1}{2}(3)\left(\cfrac{102}{5} \right)\implies A=\cfrac{1}{2}\left( \cfrac{306}{5} \right)\implies A=\cfrac{153}{5}\implies A=30\frac{3}{5}

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Match each pair of points to the equation of the line that is parallel to the line passing through the points.
Luba_88 [7]

Answer:

B(5,2) \:\: C(7,-5) \:\: \rightarrow y=-3.5x-15

D(11,6)\:\: E(5,9)\:\: \rightarrow y=-0.5x-3

H(4,4)\:\: I(8,9) \:\: \rightarrow y=1.25x+4

L(5,-7)\:\: M(4,-12)\:\: \rightarrow y=5x+9

Step-by-step explanation:

We need to match the slope of the function with the slope of the lines connecting the two points given. The slope of the lines are as follows:

B(5,2) \:\: C(7,-5)=\frac{2--5}{5-7} =-3.5

D(11,6)\:\: E(5,9)=\frac{6-9}{11-5} =-0.5

H(4,4)\:\: I(8,9)=\frac{4-9}{4-8} =1.25

L(5,-7)\:\: M(4,-12)=\frac{-7--12}{5-4} =5

F(-7,12)\:\: G(3,-8)=\frac{12--8}{-7-3}=-2

J(7,2)\:\:K(-9,8) =\frac{2-8}{7--9} =-0.375

Now,

the slope of the line BC matches with the slope of  y=-3.5x-15.

the slope of the line DE matches with the slope of y=-0.5x-3.

the slope of the line HI matches with the slope of y=1.25x+4.

the slope of the line LM matches with the slope of  y=5x+9.

and the slopes of the lines FG and JK do not match with any of the functions given.

Thus,

B(5,2) \:\: C(7,-5) \:\: \rightarrow y=-3.5x-15

D(11,6)\:\: E(5,9)\:\: \rightarrow y=-0.5x-3

H(4,4)\:\: I(8,9) \:\: \rightarrow y=1.25x+4

L(5,-7)\:\: M(4,-12)\:\: \rightarrow y=5x+9

4 0
3 years ago
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Find the value of y.
Serggg [28]
for this question you would need to utilize ratios to find the side lengths. For example, set up a ratio between the big triangle with the small triangle to find a side length
4 0
3 years ago
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I need help with my math
Tresset [83]
What is the question 
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3 years ago
If you bought a stock last year for a price of $147, and it has risen 17.9% since then, how much is the stock worth now, to the
zysi [14]

Answer:

This is an percentage question, meaning you have to calculate the percentage rise.

1) Do 147 * 17.9% (Which equals to 26.3 [1.d.p.])

2) 26.3 is the answer

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5 0
3 years ago
In ΔUVW, w = 44 cm, u = 83 cm and ∠V=141°. Find the area of ΔUVW, to the nearest square centimeter.
Tema [17]

Answer:

Area  \approx 1149\ cm^2

Step-by-step explanation:

<u>Given that:</u>

ΔUVW,

Side w = 44 cm, (It is the side opposite to \angle W)

Side u = 83 cm (It is the side opposite to \angle U)

and ∠V=141°

Please refer to the attached image with labeling of the triangle with the dimensions given.

Area of a triangle with two sides given and angle between the two sides can be formulated as:

A = \dfrac{1}{2}\times a\times b\times sinC

Where a and b are the two sides and

\angle C is the angle between the sides a and b

Here we have a = w = 44cm

b = u = 44cm

and ∠C= ∠V=141

Putting the values to find the area:

A = \dfrac{1}{2}\times 44\times 83\times sin141\\\Rightarrow A = \dfrac{1}{2} \times 3652 \times sin141\\\Rightarrow A =1826 \times 0.629\\\Rightarrow A  \approx 1149\ cm^2

So, the <em>area </em>of given triangle to the nearest square centimetre is:

Area  \approx 1149\ cm^2

8 0
3 years ago
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