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____ =  (81 * 5) / 3 =  (81/3) * 5  =   27 * 5 =   135

Let A =  (0,1)

Let B = (1,1)

Let X = ( 1, 1/2)

Let D = (0,0)

The beam will reflect from  X to D

The path length =  2 * AX  = 2 sqrt [ ( 0 -1)^2 + (1 -1/2)^2 ]  = 2 sqrt [ 1 + 1/4] =  2 sqrt (5) / 2  = sqrt 5

Angle BDX = 90 = angle ADX

DA = BD

XD = XD

So, by LL, triangles BXD and AXD are congruent right triangles

And by the same reasoning triangles, AXE and CXE  are congruent right triangles

And angle DXE  = the angle formed by the intersection of  lines p,q = 45 = the sum of  angles AXD and   AXE

But angles  BXD and CXE  = angles AXD and AXE....so their sum is  also = 45

So angle BXC = AXD + AXE + BXD + CXE    =    45 + 45  =  90

A  will have an area of (-1/5)^2   =1/25  of XYZ

So         (1/25) [ XYZ ] / [ XYZ ] =  1/25

Triangle ADC is right with angle ADC = 90

AC = 2

So DC = 2 / sqrt 2 =  sqrt 2 =  AB  = x

(2x^3 + x^2 - 3x + 5)  ( x^4 - 7x^3 + tx^2 - 11x + 15)

2x^7 - 14x^6 + 2tx^5 - 22x^4 + 30x^3

              x^6  -  7x^5  +tx^4   -   11x^3 + 15x^2

                      -  3x^5  +21x^4 -  3tx^3 +  33x^2 - 45x

                                     5x^4  +  35x^3 + 5tx^2 - 55x + 75

To  have no x^2 term

15 + 33 + 5t  =  0

48 + 5t =  0

-5t = 48

t = -48/5

sqrt ( 3^2 + 5^2 + 11^2)  = sqrt [ 155 ]  units  ≈  12.45  units

l =  21 cm

Circumference  of  cone =  (120/360)* 2pi (21) =  14 pi   cm

Radius of  cone  =   14 pi / 2pi  =   7   cm  = r

Volume of cone =   (1/3) pi r^2 * [sqrt ( l^2 - r^2 ) ]  = 

(1/3) pi (7^2) * [ sqrt ( 21^2 - 7^2) ]   ≈ 1015.94  cm^2

L =   sqrt [ 92^2  + (33-25)^2] =   sqrt [ 92^2 + 8^2] =  4sqrt [533]

R = 33

r =25

Surface area = pi L (R + r) + pi (R^2 + r^2)  = pi [ 4sqrt (533) ](58) + pi ( 33^2 + 25^2)  ≈ 22211.49 units^2

S1  =  pi * 33^2           S2  = pi * 25^2       H  = 92

Volume  =   H/3 * ( S1 + S2 +  sqrt  [ S1 * S2 ])  = 

92/3  *  ( pi (33^2 + 25^2) +  sqrt [ pi*33^2 * pi* 25^2 ] ) ≈ 244612.78 units^3

Well, THE 8x can be multiplied by 1, and the 7 can be multiplied by x so ans should be 15x

Coefficient: 15

Diagonal across the base: This can be calculated using the Pythagorean theorem on the right triangle formed by half the width, half the length, and the diagonal across the base.

Applying the formula, we get 25​2+23​2=d12​, where d1​ is the diagonal across the base. Solving for d1​, we get d1​=38​​.

Jessica has one cube with a side length of $1,$ and another cube with a side length of $1.$  She then glues the two cubes together, to form one single solid.  What is the smallest possible surface area of the single solid?  

Each cube will have a surface area of 6 square units, for a total of 12 square units.  

When Jessica sticks them together, she lost two surfaces because they're on the inside now.  

So the remaining surface area is 10 square units.  

               (AB)² • (E)   

Area  =  ––––––––––   

                    3   

The volume of a pyramid is one-third the area of the base, times the height..  

Since ABCD is a square, the area of the base is one side squared.  

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