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1.  The lateral surface area =  pi * r * L     where L is the slant height of the  cone

So

16sqrt (5) * pi = pi * 4 * L

16sqrt (5) / 4 = L  =  4sqrt (5)

Height of cone =  sqrt [ L^2 - r^2]  =  sqrt [ 80 - 4^2] = sqrt [ 64] = 8

Volume of  cone  =     pi * r^2 * height  / 3   =   pi * 4^2 * 8  / 3   =   (128/3) pi 

There are  2^5 - 1  =  31 non-empty subsets of the given one

Any subset containing "5"  is what we are looking  for

The number of non-empty subsets NOT CONTAINING "5"  =  the number of non-empty subsets of {1,2,3,4} =  2^4 -1  = 15

So....the number including "5"  =  31 - 15 =  16

I believe \(sin(\theta)\) would be \(\frac{24}{25}\).

Please correct me if i am wrong

< I can't see melody on the scoreboard >   

Click the down arrow at the bottom of the Top Users list, then look again.  

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Chris, you were right all along.  The problem stipulates x not equal to y. 

I had overlooked that in all that run-together text and those dollar signs. 

After I realized it, too late to edit, I was relieved that my comment got lost  

in the moderation by AI, which btw I maintain is a little short on the "I" part.   

Melody hadn't been answering or asking questions for the last months.

What cphil said lol

x-7≥0

x-2≥0

x<5/2 Or 5/2>x or x<13/2 or 13/2>x

Correct me if im wrong... but I'm pretty sure x can be all real numbers?

1.  (x - 5) (x + 3)  =  0

      x^2   - 2 x + - 15  = 0

3.    2x^2  -8x  + 7  =   0

Product of roots  = ab = 7/2

2ab  = 7       (1)

Sum of roots  =  a + b = - (-8) / 2 = 4

Square both sides

a^2 + 2ab + b^2 = 16      (2)

Sub (1) into (2)

a^2 + 7 + b^2   =16

a^2 + b^2  = 9

a^3 + b^3 = 

(a + b) ( a^2 - ab + b^2)  = 

(a + b) ( a^2 + b^2   - ab)  =

(4)  (9 - 7/2)  = 

(4) ( 11/2) =

22

Triangle HQN similar to Triangle FQG

HN / FG =  2 / 7

Therefore  HQ / FQ  = 2 / 7

Therefore  HQ / (HQ + FQ) =  HQ / HF  =  2 / ( 2 + 7)  = 2/9  ⇒ HQ = (2/9) HF

Triangle PFM similar to Triangle PHG

FM / HG =  3 / 8

Therefore PF / PH = 3 / 8

Therfore PF / (PF + PH)  = PF / HF  = 3 /( 3 + 8) = 3 / 11 ⇒  PF = (3/11) HF 

Getting a common denominator  between  9  and 11

HQ = 22/99FH

PF  = 27/99 FH

Therefore

FH - HQ - PF  =  PQ

FH - (22/99)FH - (27/99)FH  = PQ

(50/99)FH  = PQ

Therefore

PQ / FH  =  50 / 99

Can you post the picture by any chance?

We call one base x.

\(\frac{(x+x+4)12}{2} = 80\)

\((x+x+4)6 = 80\)

\(x+x+4 = 40/3\)

\(2x = 28/3\)

\(x = 14/3\)

\(x+4 = 26/3\)

These are the side lengths for the 2 bases. I do not know the other 2 lengths, since i do not know where they are positioned.

Perhaps you have a picture you couldn't post, so I suggest using the pythagorean theorem for the other 2 sides, one side being the height.

The formula for the area of a Trapezoid is : A=( a+b )h/2 (a+b are the numerators)

a and b are the bases, so we can say a=6 and b=22

h=4 (height)

A= (6+22)4 / 2

A= 28*4/2

A=28 * 2 

A=56in²

if im incorrect someone please correct me!

Dang, it was wrong

It was correct, thanks

This one is pretty straightforward, you should try it.

\(3a^5-a^4+4a^3+22a^2-4a+48\)

Alrighty...

3. For this one, I will calculate the white parts and subtract from the volume. You can try and do it the other way around.

The pyramid ABCF has an area that is half of half of the rectangular prism. You can see the divisions, FC and triangle ACF.

Same thing for the other white parts, so this sums to 3/4 of the total volume, therefore the shaded area is 1/4 of the prism. The volume is 60.

1/a  +   1/b =    (a + b)   /  (ab)

x^2 - 22x  - 9  =   0 

Sum of roots  = a + b =  - (-22) / 1  = 22

Product of roots  = ab  =  -9/1 =  -9

So

(a + b) / (ab) =   22 /  -9

(x  - 5) ( x + 5)  =  0 

x^2   - 25  =  0

x^2  +  0x  +   -  25  =   0

Rearrange as

x^2  + 26  =  0

x^2  =  -26              take  both roots

u =  i sqrt (26)              and      v  =  -i sqrt (26)

u / v  +   v /  u    =

i sqrt (26) / -i sqrt (26)    +     -i sqrt (26)  / i sqrt (26)  = 

- 1        +      -1    =

-2

Yaey  I got it right! Currently reading number 3.

The numbers \(x_1,x_2,x_3,x_4\) are chosen at random in the interval \([0,1]\)  Let \(I\) be the interval between \(x_1,x_2\) and let \(J\) be the interval between \(x_3,x_4\)  Find the probability that intervals \(I,J\) both have length greater than \(\frac{3}{4}\). 

Ok, so this implies that \(x_2-x_1>3/4\), same thing for the other pair.

We'll just assume that the numbers go from least to greatest.

This means that \(x_1<1/4\), which is 1/4 chance, times the chance that \(x_2>3/4\), which is also 1/4. multiplying, we get 1/16, which we multiplu by 2 since they don't have to go from least to greatest, we just assumed. This gives us 1/8.

We do the same for \(x_3,x_4\) and get the same result. Adding these together will give us our final answer, \(\fbox{1/4}\)