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lara [203]
4 years ago
12

find the surface area of a right regular hexagonal pyramid with sides 3 cm and slant heights 6cm. show all of your work.​

Mathematics
1 answer:
DedPeter [7]4 years ago
6 0

Answer:

The surface area of right regular hexagonal pyramid = 82.222 cm³

Step-by-step explanation:

Given as , for regular hexagonal pyramid :

The of base side = 3 cm

The slant heights = 6 cm

Now ,

The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times a^{2} + 3 a \sqrt{h^{2}+ 3\times \frac{a^{2}}{4}}

Where a is the base side

And h is the slant height

So, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 3^{2} + 3 \times 3 \sqrt{6^{2}+ 3\times \frac{3^{2}}{4}}

Or, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 9 + 9 \sqrt{36+ 3\times \frac{9}{4}}

Or,  The surface area of right regular hexagonal pyramid = 23.38 + 9 × \frac{171}{4}

∴  The surface area of right regular hexagonal pyramid = 23.38 + 9 × 6.538

I.e The surface area of right regular hexagonal pyramid = 23.38 + 58.842

So,  The surface area of right regular hexagonal pyramid = 82.222 cm³ Answer

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Answer:

5(9x + 2y + 10)

or

x = 0

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Step-by-step explanation:

If you are just factoring this, you will factor a 5 out from everything.

5(9x + 2y + 10)

or if you are trying to solve just isolate one of the variables and then substitute it into the original equation

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10y = 50 - 45x

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45x + 10(5-4.5x) + 50

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The location of point V is (-3,3). The location of point X is (9,13). Determine the location of point W which is 3/4 of the way
Inga [223]

ANSWER

W(\frac{9}{7},\frac{51}{7} )

EXPLANATION

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The x-coordinate of this point is given by:

x= \frac{mx_2+nx_1}{m + n}

x= \frac{3(9)+4( - 3)}{4 + 3}

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The y-coordinates is given by;

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W(\frac{9}{7},\frac{51}{7} )

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