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The length of a rectangle is 51 inches more than the width. The perimeter is 146 inches. Find the length and the width.

146 - 102 = 44       44 / 4 = 11

The width is  11 in

The length is    (146 - 22) / 2 = 62 in  

Assuming \(2^{x^2} \) means \(2^{(x^2)}\) then:

Triangle ABC is an equilateral triangle and O is the center of its inscribed circle. If the area of the circle is  4*pi cm², what is the area, in square centimeters, of triangle ABC? Express your answer in simplest radical form.

The area of  Δ ABC ≈ 20.785 cm²    
 

Let n = initial number of stamps

We are told:   n + 31 = 3*(n - 31)

I'll leave you to determine n from this and hence the number, (n + 31), that Cheko ended up with.

Probability = (10/50)*(9/49)*(8/48)*(8/47)*(7/46) ≈ 0.000159

This graph should help:

I assume this means that the ice cream is half in and half out of the cone, in which case:

The diameter of the cone is 6 inches (big cone!) as circumference = π*diameter.

Hence bottom of cone is 3 inches below cone opening.

Hence depth of cone = 3 + (1/3)*3 = 4 inches.

Volume of cone, Vc = (1/3)π*3² *4 in³

Volume of ice-cream hemisphere inside cone, Vi = (1/2)*(4/3)*π*3³ in³.

So volume of cone not filled with ice cream V = Vc - Vi

If  f(n) = 3f(n-2) - 2f(n-1)   and

f(1) = -1

f(2) = 3

We can simply continue:

f(3) = 3f(3-2) - 2f(3-1) = 3*(-1) - 2*3 = -9

f(4) = 3f(4-2) - 2f(4-1) = 3*3 - 2*(-9) = 27

f(5) = 3f(5-2) - 2f(5-1) = 3*(-9) - 2*27 = -81

More generally:

f(n) = (-1)ⁿ*3^(n-1)

Hi Mellie,

Have you tried my method?  I never have any problems. 

You should not be cheating on your homework.

Can you confirm whether you intend

\({2^{x^{2}}}={2^{(x^{2})}}\quad or \quad {(2^{x}})^{2}\)

I think if your notation is correct it should be the first one. 

That is the one I have been playing with.  But so far I am baffled.

1 post= 1 question

nvm the answer is 18 in.^2

No Chris is welcome to the lead    

Let the celebrity death match begin !!!      haha      ~EP

w = width

L= length = 51 + w        given

perimeter =   w + w + l+ l =     2 w    + 2(51+w)   = 146      solve for w, then l = w+ 51

Nice Job Chris!

Please do not post AoPS problems on this site.

Thank you.

Do you mean the sum total of all 6 dice, ranging from sum of 6 to sum of 36. If so, you have the following probabilities:

6 - 1 / 46656,
7 - 1 / 7776, 
8 - 7 / 15552, 
9 - 7 / 5832, 
10 - 7 / 2592, 
11 - 7 / 1296, 
12 - 19 / 1944, 
13 - 7 / 432, 
14 - 43 / 1728, 
15 - 833 / 23328, 
16 - 749 / 15552, 
17 - 119 / 1944, 
18 - 3431 / 46656, 
19 - 217 / 2592, 
20 - 469 / 5184, 
21 - 361 / 3888, 
22 - 469 / 5184, 
23 - 217 / 2592, 
24 - 3431 / 46656, 
25 - 119 / 1944, 
26 - 749 / 15552, 
27 - 833 / 23328, 
28 - 43 / 1728, 
29 - 7 / 432, 
30 - 19 / 1944, 
31 - 7 / 1296, 
32 - 7 / 2592, 
33 - 7 / 5832, 
34 - 7 / 15552, 
35 - 1 / 7776, 
36 - 1 / 46656

Can you show us the working you have done yourself?

Maybe upload a photo of a sketch.

Thanks a lot man!

The shortest distance from A to B that can be traced on the surface is the sum of longest diagonal of the block's surface (26cm) and the thickness of the block ( 6cm)

The length of diagonal  D = sqrt ( 24² + 10² ) = 26 cm

The shortest distance from A to B that can be traced is:      26 + 6 = 32 cm