geno3141

The front side is a triangle:  A  =  ½·base·height  =  ½·20·21  =

The back side is identical to the front:  A  =

The base is a rectangle:  A  =  20 x 22  =

The slanted side is a rectangle:  A  =  22 x 29  = 

The left side is a rectangle:  A  =  21 x 22  =

You'll need to add these answers together ...

Yes.

Yes.

Yes.

For the function  f(n)  =  2 + 2(n + 1)  =  2 + 2n + 2  =  2n + 4

the oefficient of the variable is  2  and the constant is  4.

Since it is an equilateral triangle, each of the interior angles is 60^o.

Look at the large triangle  -- from the center of the circle to the point of tangency to the exterior vertex.

The angle formed by the radius and the tangent is 90^o.

The angle at the center of the circle is 60^o.

Therefore, angle(x) is 30^o.

Using the distance formula:  D  =  sqrt[ (x₂ - x₁)² + (y₂ - y₁)² ]

A to B:  sqrt[ (2 - 0)² + (5 - 1)² ]  =  sqrt[ (2)² + (4)² ]  =  sqrt( 20 )  =  2sqrt(5)

B to C:  sqrt[ (5 - 2)² + (2 - 5)² ]  =  sqrt[ (3)²+ (-3)² ]  =  sqrt( 18 )  =  3sqrt(2)

C to D:  sqrt[ (7 - 5)² + (0 - 2)² ]  =  sqrt[ (2)² + (-2)² ]  =  sqrt( 8 )  =  2sqrt(2)

D to A:  sqrt[ (0 - 7)² + (1 - 0)² ]  =  sqrt[ (-7)² + (1)² ]  =  sqrt( 50 )  =  5sqrt(2)

Can you add these distances together?

Let  x,  x + 2,  and  x + 4  represent the three consecutive positive integers.

The product of the smaller two:  x(x + 2)

3 more than four times the sum of the larger two:  3 + 4[ (x + 2) + (x + 4) ]

       x(x + 2)  =  3 + 4[ (x + 2) + (x + 4) ]

        x² + 2x  =  3 + 4[ 2x + 6 ]

        x² + 2x  =  3 + 8x + 24

        x² + 2x  = 27 + 8x

  x² - 6x - 27  =  0

(x - 9)(x + 3)  =  0

x  =  9

x + 2  =  11

x + 4  =  13

z  = -3 + 3i

To find the value of r:  r  =  sqrt( (-3)² + (3)² )  =  sqrt( 9 + 9 )  =  sqrt(18)  =  3sqrt(2)

To find the angle:  tan^-1( 3 / -3 )  =  tan^-1( -1 )  =  3pi / 4       [remember that it is in the second quadrant]

3sqrt(2) cis( 3pi/4)    

To find the three cube roots of  -2 - 2i

find the radius:  sqrt( (-2)² + (-2)² )  =  sqrt( 4 + 4 )  =  sqrt(8)

and the angle:  tan^-1( -2 / -2 )  =  5pi/4     [remember that it is in the third quadrant]

We now have:  8^1/2cis( 5pi / 4 )

To find the first cube root, we need to find the cube root of  8^1/2  which is  (8^1/2)^1/3  =  8^1/6                            

and to divide the angle by 3, giving us  5pi /12.

So, for our first answer, we have  8^1/6 cis ( 5pi / 12 )

To find the next answer, we need to add 1/3 (or 2pi / 3 ) of the circle to our previous answer:   8^1/6 cis ( 13pi / 12 )

and to get our last answer we need to add another  2pi / 3 to this answer:   8^1/6 cis ( 21pi / 12 )  =  8^1/6 cis ( 7pi / 4 )

For the second problem:

   24 | 5

   25 | 0, 1, 3, 8

   26 | 4, 7, 9

   27 | 3, 6

For the first problem: 

   0 | 3, 8

   1 | 2, 2, 4

   2 | 0, 1, 3, 6

   3 | 4