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Helga [31]
3 years ago
6

The sides of a hexagon are 2, 3, 2, 4, 7, and 6. Find the perimeter of a similar hexagon with two sides of length 3.

Mathematics
2 answers:
Arturiano [62]3 years ago
8 0

Answer:

36

Step-by-step explanation:

The previous answerer forgot to add the two 3's to the sum so the correct answer it 36

Basile [38]3 years ago
6 0

The first hexagon: 2, 2, 3, 4, 6, 7

The second hexagon: 3, 3, a, b, c, d

The hexagons are similar. Therefore the sides are in proportion.

\dfrac{a}{3}=\dfrac{3}{2}     <em>multiply both sides by 3</em>

a=\dfrac{3}{2}=4.5

\dfrac{b}{4}=\dfrac{3}{2}    <em>multiply both sides by 4</em>

b=\dfrac{3}{2}\cdot4=6

\dfrac{c}{6}=\dfrac{3}{2}     <em>multiply both sides by 6</em>

c=\dfrac{3}{2}\cdot6=9

\dfrac{d}{7}=\dfrac{3}{2}    <em>multiply both sides by 7</em>

d=\dfrac{21}{2}=10.5

The perimeter:

P=4.5+6+9+10.5=30

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natali 33 [55]

Answer:

6

Step-by-step explanation:

6 * 3 = 18

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4 0
3 years ago
A preimage is the the original figure in a transformation.<br> A) True B) False
Kitty [74]
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33. Suppose you were on a planet where the
tatyana61 [14]

<u>Answer:</u>

a) 3.675 m  

b) 3.67m

<u>Explanation:</u>

We are given acceleration due to gravity on earth =9.8ms^-2

And on planet given = 2.0ms^-2

A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula  </u>

\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}

Where H = max jump height,  

v0 = velocity of jump,  

Ø = angle of jump and  

g = acceleration due to gravity

Considering velocity and angle in both cases  

\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)  

g1 = 2.0ms^-2 and  

g2 = 9.8ms^-2

Substituting these values we get H1 = 3.675m which is the required answer

B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by  </u>

 \mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}

which is due to projectile motion of ball  

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,  

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write  

\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0ms^-2  and g2 = 9.8ms^-2

We get h1 = 3.67m which is the required height

6 0
3 years ago
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Vikentia [17]
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