CPhill

2.  The diagram is not drawn to scale, but the measurements of the line segments and the right angles are correctly labeled. As shown in the diagram, BE=EC. Are the red and blue triangles similar? If so, enter the side of the blue triangle that corresponds to AB. If not, enter "no."

It's obvious that, if the triangles are similar, then ΔABE is similar to ΔECD

Note that....if similar, the scale factor is  120/35 = 24/7

Since...if similar, then DC =  (7/24)BE = (7/24)CE

Let CE = x   ....so DC = (7/24)x

Then, by the Pythagorean Theorem,

sqrt ( x^2 + (7x/24)^2 ] = 35

x *sqrt [ 576 + 49] / 24  =  35

x * sqrt (625) = 24 * 35

x * 25 =  840

x = 840 /25 =  33.6  =  CE  = BE

And...if similar..... AB = (24/7)CE

So....by the Pythagorean Theorem......

sqrt (BE^2 + AB^2 ) = 

sqrt (BE^2  + (24 CE/ 7)^2 ) =

sqrt ( 33.6^2 + [24 (33.6)/ 7]^2 )  =

sqrt ( 1128.96 + 13.271.04) =

sqrt ( 14400) =  120 = AE

So.....they are similar

AB  is similar to EC

3/7 x 1/2 =   3/14

(3/5 - 1/3) =  [ 9 - 5 ] / 15  =   4 / 15

So

(3/14)   /    (4/15)   =

(3/14) * ( 15/4)  =

45/ 46

The first one is answered, here :

https://web2.0calc.com/questions/plz-help-and-explain-thoroughly-helpp

4. We know Angle A=45 degrees, Angle B=15 degrees and BC=sqrt(6). What is AB?

Angle C =  120°

Using the Law of Sines

BC / sin A   =  AB/ sin C

sqrt (6) /  (1/sqrt(2) ) =  AB / ( sqrt (3) / 2)

sqrt (12) = 2 AB / sqrt (3)

sqrt (3) * sqrt (12) = 2 AB

sqrt (36)  = 2 AB

6  =  2AB

AB =  3

No big deal.....easy to make mistakes....!!!!

We have the following number of possible sets of numbers.....we have six numbers and we want to choose any two of them....so....

C(6,2) = 15 sets

These are 

(1,2) (1,3) (1,4) (1, 5) (1,6)

(2, 3) (2,4) (2, 5) (2, 6)

(3,4) (3,5) (3,6)

(4, 5) ( 4, 6)

(5, 6)

Note, Ant, that only two of them have a product of six

So.....thirteen don't

We had a robot check that.....he/she found no problem.....apparently, robots aren't expert with the finer details of the English language.....

[ We should have one of Ginger Ale's chimps check this.....they rarely miss anything....particularly after a few stiff drinks... ]

Let r be the radius of the  tank

The volume of the tank is

6000pi = pi * r^2 * h

Solving for h, we have that

h = 6000/r^2

So....the surface area cost is given by

Sa cost  =   ($1)pi*r^2 + ($2)pi *r^2 + ($4)2pi * r * 6000/r^2

Sa cost  =  3pi*r^2 + 8pi* 6000r^(-1)

Take the derivative of this to find the radius that minimizes the cost

Sa ' =   6pir - 48000pi / r^2

Set this to 0  and solve for r

pi ( 6r - 48000/r^2)  = 0

6r -  48000/^2  = 0

6 ( r - 8000/^2) = 0

r - 8000/r^2 = 0

r^3 - 8000  = 0

r^3 = 8000

r = 20 ft

So....the minimum cost is 

3pi*(20)^2 + 8pi* 6000/20 =

pi [ 1200 + 2400]  =   $ 3600pi    ≈  $11309.73

( f + g) (x)  = 

 f(x) + g(x) =

[ x^3 - x ] + [ x^3 + 2x^2 - 10 ]  =

2x^3 + 2x^2 - x - 10

( f - g) (x) =

f(x) - g(x)   =

[ x^3 - 2x^2  + 12x - 6 ] - [ 4x^2 - 6x + 4 ]  =

x^3 - 6x^2 + 18x - 10

( f * g)(x)  =

f(x) * g(x) =

[x^2 + 2x - 6 ] * [ x^4  + 3 ]   =     

x^6 + 2x^5  - 6x^4 + 3x^2 + 6x - 18  

No prob.....!!!