Given: δwxy is isosceles with legs wx and wy; δwvz is isosceles with legs wv and wz. prove: δwxy ~ δwvz

By AA

ΔWXY ~ΔWVZ

Step-by-step explanation:

Here WXY is an isosceles triangle with legs WX & WY

So WX = WY

Hence ∠X = ∠Y

So ∠2= ∠3.

Now by angle sum property

∠1 + ∠2+∠3 = 180°

∠1+∠2+∠2=180°

2∠2 = 180° - ∠1     .......(1)

In triangle WVZ

WV = WZ

So ∠V = ∠Z

∠4 = ∠5

Once again by angle sum property

∠1 + ∠4 + ∠5=180°

∠1 + ∠4 + ∠4 = 180°

2∠4 = 180° - ∠1       ...(2)

From (1) & (2)

2∠2 = 2∠4

∠2=∠4

Now ∠W is common to both triangles

Hence by AA

ΔWXY ~ΔWVZ

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}