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Sorry.....my answer should have been in cubic feet........

We need to find  the height of the pyramid.....this is given by :

tan (35)  =  H / [(3/2)sqrt(3)]      divide both sides by [(3/2)sqrt(3)]

tan (35) * [(3/2)sqrt(3)]  = H   

And the volume is given by :

(1/3) * Area of the base * height

And the Area of the base = 6 (1/2)*3^2* sin(60)  =  27*sin(60)

So....the volume is :

(1/3) * 27 * sin (60) * tan (35) * [(3/2)sqrt(3)]  = about 14.18 units^3

A "quadrilateral" is just a shape with 4 sides.

Can a four sided shape have 2 right angles and 2 parallel sides?

Does CPhill's drawing have 4 sides?

Does it have 2 right angles?

Does it have 2 parallel sides?

If it passes all those tests, then there is an example of a four sided shape with 2 right angles and 2 parallel sides.

calculate how many lbs of sulfur are contained in 3,805 lbs of coal that have 5%of pure sulfur.

Just multiply:

5% x 3,805 =5/100 x 3,805 =190.25 pounds of pure sulfur.

So NO?

Or yes?

I think its no

See what you think :

AB is parallel to CD.....angle BAC  and angle DCA  are right

You might also like to investigate Kepler's 3 Laws of Planetary motions:

Johannes Kepler, working with data painstakingly collected by Tycho Brahe without the aid of a telescope, developed three laws which described the motion of the planets across the sky.

1. The Law of Orbits: All planets move in elliptical orbits, with the sun at one focus.

2. The Law of Areas: A line that connects a planet to the sun sweeps out equal areas in equal times.

3. The Law of Periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.

Kepler's laws were derived for orbits around the sun, but they apply to satellite orbits as well.

2^(3x-2)=3^(2x-1)      take the log of both sides

log 2^ (3x - 2)  =  log 3^(3x - 1)   and we can write

(3x - 2) log 2  =  (3x - 1) log 3  simplify

3x* log 2  - 2 log 2  =  3x log 3  - log 3   rearrange as

3log 2 * x - 3 log 3 * x   =  2log 2 - log 3       implify the left side

x [ 3 log 2  - 3 log 3 ]   = 2 log 2 - log 3       divide both sides by  [ 3 log 2  - 3 log 3 ]

x = [ 2 log 2 -  log 3 [ / [3 log 2 - 3 log 3 ]   ≈ -0.2365

You do not use gravity at all.

You may use the radius of the earth depending on the question.

Hi Sara :)

Lets see :)

\(\sqrt3\;x^2-2\sqrt2\;x-2\sqrt3=0\\~\\ \triangle=(-2\sqrt2)^2-4*\sqrt3* -2\sqrt3 \\ \triangle=(8+24) \\ \triangle=32\qquad \text{Positive so this means it has 2 real roots} \\~\\ \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-b \pm \sqrt{\triangle} \over 2a}\\ x = {2\sqrt2 \pm \sqrt{32} \over 2\sqrt3}\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\\ \\~\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\times \frac{\sqrt3}{\sqrt3}\\ x = {2\sqrt6 \pm 4\sqrt{6} \over 6}\\ x = {\sqrt6 \pm 2\sqrt{6} \over 3}\\ x = {\sqrt6 (1\pm2) \over 3}\\ x=\frac{3\sqrt6}{3}\qquad or \qquad x=\frac{-\sqrt6}{3}\\ x=\sqrt6\qquad or \qquad x=\frac{-\sqrt6}{3}\\ \)

I checked this answer by graphing it and it is correct.

The best way is to focus. Don't stress about answers you don't know and skip questions you can't solve and come back to them later. Any person who would ask that question has already aces the test.

Also, you can simplify √27:

√27 = √9•3 = 3√3

Wiat never mind it is showing now!

√h³ = √h•h•h = (√h•h)(√h) = h√h

√8 = √4•2 = (√4)(√2) = 2√2

final result is h√h over 2√2

Hi MJ

If it is a proper question then type it properly.

Or are you just bored?

Yea ok April Fool's Day - You got me.  Trouble is you are late!  

In Sydney it is 7am   the Second  of April      

i is the square root of -1, so i squared is -1.

I dnot un,dderstand...

#catface

#whocares

#kingdomofcats

#idontknowtheanswer

can you explain why

I just figured out how to use LaTeX! So it's:

\(\frac{\sqrt{27}}{m^{2}*\sqrt{m}}\)

How could you square an eye? 

sqrt(27)/sqrt(m²) = sqrt(27)/m²*sqrt(m)

basically: Square root of 27 divided by m squared times the square root of m