hectictar

YZ  is formed by the angle marked by one red line.

JG  is formed by the angle marked by no red lines.

So we cannot say that YZ = JG .

XY is formed by the angle marked by two red lines.

JH  is formed by the angle marked by one red line.

So we cannot say that  XY = JH

∠Y is marked by no red lines.

∠J  is marked by two red lines.

So we cannot say that ∠Y = ∠J

∠Y is marked by no red lines.

∠H is marked by no red lines.

So ∠Y = ∠H

the number of teams after round  0   =   128

the number of teams after round  1   =   (1/2) * 128

the number of teams after round  2   =   (1/2) * (1/2) * 128

the number of teams after round  x   =    (1/2)^x * 128

(1/2)^x * 128   =   16

                                     Divide both sides of the equation by  128 .

(1/2)^x   =   1/8

                                     We can express  1/8  as  (1/2)³

(1/2)^x   =   (1/2)³

                                     So we can see that...

x  =  3

Wait a minute!!!

A. Trapezoid EFGH is not congruent to trapezoid E′F′G′H′ because there is no sequence of rigid motions that maps trapezoid EFGH to trapezoid E′F′G′H′ .

Here's how I like to think of these...

original perimeter   =   2(L + W)

To scale the photo, mulitply the length by  1.75  and the width by  1.75 , so that

the new length  =  1.75L   and   the new width  =  1.75W

new perimeter   =   2(1.75L + 1.75W)

new perimeter   =   1.75 * 2(L + W)

new perimeter   =   1.75 * original perimeter

new perimiter    is   1.75 times the original perimeter

original area   =   L * W

new area   =   1.75L * 1.75W

new area   =   1.75 * 1.75 * L * W

new area   =   1.75² * L * W

new area   =   1.75² * original area

new area   =   3.0625 * original area

new area   is   3.0625 times the original area

8 m  increased by  25%  of  8 m

=  8 m  +  25%  of  8 m

=  8 m  +  0.25 * 8 m

=  1.25 *  8 m

=  10 m

a)  We can use the law of cosines to find the angles.

c²   =   a² + b² - 2ab cos C    , where  C  is the angle opposite side  c .

Let  a = 19 ,  b = 20 ,  c = 21  and  A ,  B , and  C  be the angles opposite their respective sides.

21²   =   19² + 20² - 2(19)(20)cos C

441   =   361 + 400 - 760 cos C

441   =   761 - 760 cos C

                                                     Subtract  761  from both sides of the equation.

-320   =   -760 cos C

                                                     Divide both sides by  -760 .

8/19   =   cos C

                                                     Take the inverse cosine of both sides.

C  =   acos( 8/19 )   ≈   65.099°

Now let's use the law of cosines again to find the angle opposite the  20 cm  side.

20²   =   19² + 21² - 2(19)(21)cos B

400   =   361 + 441 - 798 cos B

400   =   802 - 798 cos B

-402   =   -798 cos B

67/133   =   cos B

B   =   acos( 67/133 )   ≈   59.751°

Since there are  180°  in every triangle,  A + B + C  =   180°   and   A  =  180° - B - C

A   ≈   180° - 59.751° - 65.099°   ≈   55.15°

b)

Let the base be the 19 cm side, and the height be  20 sin 65.099°

area   ≈   (1/2)(19)(20 sin 65.099°)   ≈   172.337  sq cm

They are alternate interior angles.

7.   a/x = c/a     ~     Corresponding parts of similar triangles are proportional

11.   ΔCBA ~ ΔDCA     ~     AA Similarity Postulate

12.   b/y = c/b    ~     Corresponding parts of similar triangles are proportional

15.   a^2 + b^2  =  c(x + y)     ~     Distributive Property

I think that is it.....

I think they are correct!

"log" is short for "logarithm" .