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Alinara [238K]
3 years ago
6

The stained glass window is a regular hexagon which is 2.5 meters on a side. What is the approximate length of the metal frame s

urrounding the window.
Mathematics
2 answers:
e-lub [12.9K]3 years ago
8 0

Answer:

15 m

Step-by-step explanation:

A regular hexagon has all 6 sides equal in length

Length of the frame = perimeter

= 6 × 2.5

= 15 m

posledela3 years ago
3 0

Answer:

15 meters

Step-by-step explanation:

One side is 2.5 meters, so if they're all the same then you would multiply that by 6

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<span>avg = (1 + n) / 2</span> 
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3 years ago
(2sin60°)(3tan60°)—(4tan45°)(2cos45°)​
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Answer:

9 - 4\sqrt{2}

Step-by-step explanation:

sin60° = \frac{\sqrt{3} }{2}

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(2sin60°)(3tan60°) - (4tan45°)(2cos45°)

now substitute the values of each trigonometrical angles

(2*\frac{\sqrt{3} }{2} )(3*\sqrt{3} ) - (4*1)(2*\frac{1}{\sqrt{2} } )

\sqrt{3}  * (3\sqrt{3}) - 4*(\frac{2}{\sqrt{2} } )

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9 - 4 * \sqrt{2}

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5 0
3 years ago
jason mixed 8ml of a salt solution with 11ml of a 47% salt solution to make a 43%salt solution. find the percent salt concentrat
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Let x% be the salt concentration of first solution.

Then in 8ml, amount of salt = \frac{8x}{100}

Given that to this 8ml solution, 11ml solution of a 47% salt solution is mixed.

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So, total amount of salt in new mixture = \frac{8x}{100} +\frac{517}{100} = \frac{8x+517}{100}

But given that this new solution is a 43% salt solution.

Hence amount of salt in new mixture of (8+11)=19ml solution of 43% salt solution is \frac{19*43}{100} = \frac{817}{100}

Hence \frac{8x+517}{100} = \frac{817}{100}

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Hence the salt concentration in the first solution is 37.5%

8 0
3 years ago
Find the value of x, y, and z in the rhombus below.
Yuki888 [10]

Solution:

<u>It should be noted:</u>

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<u>Thus:</u>

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<u>Finding x:</u>

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4 0
2 years ago
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gulaghasi [49]

Answer:

y=t−1+ce

−t

where t=tanx.

Given, cos

2

x

dx

dy

+y=tanx

⇒

dx

dy

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2

x=tanxsec

2

x ....(1)

Here P=sec

2

x⇒∫PdP=∫sec

2

xdx=tanx

∴I.F.=e

tanx

Multiplying (1) by I.F. we get

e

tanx

dx

dy

+e

tanx

ysec

2

x=e

tanx

tanxsec

2

x

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ye

tanx

=∫e

tanx

.tanxsec

2

xdx

Put tanx=t⇒sec

2

xdx=dt

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t

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(t−1)+c

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