CPhill

The only  difference is the volume of a cylinder  with  a radius and height of 2 cm  and the volume of a hemisphere with a radius of 2cm

It obvious that the cylinder will have a greater volume than the hemisphere.........and this difference is

pi (2)^2 * 2  -  (4/6)pi (2)^3   = pi [ 8 - 16/3]  = 8pi/3   cubic cm

So...the hemisphere uses  [8pi/3] cm^3   less ice cream

2/3 of 3/4   =  6/12  = 1/2 

There may be other methods of solving this one.....but here's my attempt.....

Locate  point D  on BC and let it lie x units from the origin......and let a be the side of the square

Locate point E on  AC  and  F on AB  such that  DE = DF  = a   and DE,DF are perpendicular to each other

Now angle ACB  = angle FDB = atan(3/4)

 So, cos [atan(3/4)] = DB/ FD = x / a    →  x = ( a)cos[ atan(3/4)]

And sin ACB = sin [atan (3/4)]

So, sin [atan (3/4) ]  = a / [ 4 - x]

And substituting for x, we have

sin [atan(3/4)] = a / [ 4 - a* cos [atan(3/4)]]

And with a little help from WolframAlpha,  a = 60/37.........and this is the side  of the square

And x = BD = (60/37)cos[atan(3/4)] = 48/37

Here's the [ approximate] pic :

If you take trig......you will run across a "formula"  for the area of a sector of a circle

It's given by :

A = (1/2) r^2 (theta)      where "theta"  is the  "central angle"  of the sector measured  in radians  ....[ don't worry about the meaning of all these terms....... it isn't critical, here ]

The "central angle" measure of a sector formed by the whole circle  =  2*pi   radians

Thus,  the area  of the whole circle  =

(1/2) r^2  [ 2 * pi]  =

(1/2)(2)* pi * r^2  =

pi * r^2

Which is the familiar "formula" for the area of a circle......!!!!

Thus "pi" helps us find this area........

BTW.....before the "exact" value of "pi" was known [ not actually possible], mathematicians from India sometimes  used the value sqrt(10)  = about  3.16

The %error  here  is     [3.16 - 3.14] / 3.14 = about  0 .636%   [less than 1 percent !!!  ]

I believe this has something to do with the angle of incidence and reflection.

The angle of elevation formed by her line of sight and her distance from the mirror  is given by :

atan [5 / 3]   = about 59.04°

And this would be the same angle of elevation formed by the distance the mirror is from the top of the post  and its distance to the post....so we have......

tan (59.04)  = h / 6      where h is the post height

[5/3]  = h / 6     cross-multipy

5 * 6 / 3 = h     =  30/ 3  = 10 ft tall

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I'd like to have someone else look at this one......I might be out in left field, here....!!!!!

Here's a slightly easier method that doesn't require a calculator

27^x = 9^4

[3 ^3]^x  = [3^2]^4

[3]^(3x)  = [3]^(80        and since the bases are equal, we can solve for the wxponents

3x = 8     divide both sides by 3

x = 8/3   =  2+ 2/3   =  2.6666......   [6's repeat ]

>   = greater than

<   = less than

Here's a good way to remember this.......think of them as arrowheads........and the arrowheads always point to the smaller number......!!!!!

Are you serious?????.....It would take us 5 hours to answer all of these.....we're here to help.....NOT to do your homework for you!!!!!!

We can use the tangent function here......let h be the height of the cliff  and d be the initial distance from the boat to the base of the cliff

And we have, in the intial observaation:

tan 44  = h / d   →    h  = d tan 44

And, in the second observation, we have

tan 38  = h / [ d + 200 ]    =   [ d tan 44]  / [ d  + 200]     multiply both sides  by [ d + 200]

[d + 200] tan 38  = d tan 44

d tan 38  + 200 tan 38  = d tan 44       rearrange

d tan 38 - d tan 44  = - 200 tan 38     factor out d on the left

d [ tan 38 -  tan 44]  = - 200 tan 38       divide both sides by [ tan 38 - tan 44]

d = -200 tan38 / [ tan 38 - tan 44]   =   about 847.4 m

And the heght of the cliff =  (847.4) tan (44)  = about 818.3 m

2^x + 2^(-x)  = 6  rewrite as

2^x +  1 / [2^x)  = 6        multiply through by  2^x

[2^x][2^x] + 1  = 6[2^x]    subtract 6[2^x]  from both sides

[2^x]^2 -+ 1 - 6[2^x]   = 0       let 2^x = a   ..... and we have.....

a^2  + 1 - 6a  = 0     rearrange

a^2 - 6a + 1 = 0

Using the quadratic formula to solve for a, we have  .......a = 3-2 sqrt(2)   or  a = 3 + 2sqrt(2)

Thus  a = 2^x

And we have that

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

log2^x   = log [3 -2 sqrt(2) ]     and by a log property, we can write

x*log(2)  = log[ 3 - 2sqrt(2)]   divide both sides by log(2)

x = log[3 - 2sqrt(2)] / log(2)  = about -2.5431

In similar fashion, x also equals.....

 log[ 3 + 2sqrt(2)] / log(2)  = about 2.5431

Here's a graph of  both sides of this strange equation  with the intersection points being the solutions.....

https://www.desmos.com/calculator/uomirk7pas