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3241004551 [841]

3 years ago

8

What is the area of this regular hexagon that has been divided into six congruent triangles?

2 answers:

Softa [21]3 years ago

8 0

The answer would be 336cm^2

Maslowich3 years ago

4 0

336 cm2 is the answer

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Laura's uncle donated 120 cans of juice and 90 packs of cheese crackers for the school picnic. Each student is to receive the sa

liubo4ka [24]

Ok, First you have to find the GCF of 120 and 90. When I got the answer I got a GCF of 30. I got 30 by multiplying all the number prime numbers 120 and 90 had, 2×3×5=30 What I did next was divide thirty by the number of cracker and also by the number of can of juice.

120÷30= 4

90÷30= 3

So as you can see there could be a maximum 30 of kid's. Each of the kids would receive 4 cans of juice and 3 packs of cheese crackers.

I hope you enjoyed my explanation. :)

3 0

3 years ago

Read 2 more answers

Name the shortest segment from Point A to Line CD. Write and Solve an inequality for X.

Misha Larkins [42]

Attached the solution with work shown.

5 0

3 years ago

Which expression is equivalent to 5(w+4) when w=9

erica [24]

Answer: Should be 5(9)+4

Step-by-step explanation:

8 0

3 years ago

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In the game Kazoo, teams lose one point each time they skip a turn. They earn one point each time their team guesses correctly.

enot [183]

If they have -8 points and +6 points, that'd leave them with a total of -2 points.
So the answer is -2.

4 0

2 years ago

If tan theta =3, find the value of tan theta + tan (theta+pi) + tan (theta +2pi)

sweet-ann [11.9K]

Answer:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

Step-by-step explanation:

Given

$tan\theta = 3$

Required

Find

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$

Calculate $tan(\theta + \pi) \ and\ tan(\theta+2\pi)$

Using tan rule

$tan(\theta + \pi) = \frac{tan\theta + tan\pi}{1 - tan\theta tan\pi}$

So:

$tan(\theta + \pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + \pi) = \frac{tan\theta}{1 }$

$tan(\theta + \pi) = \tan\theta$

$tan(\theta+2\pi)$

$tan(\theta + 2\pi) = \frac{tan\theta + tan2\pi}{1 - tan\theta tan2\pi}$

$tan(\theta + 2\pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + 2\pi) = \tan\theta$ '

'So:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$ $= tan\theta +tan\theta +tan\theta$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 3+3+3$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

4 0

2 years ago