Can someone help me solve (a) and (c) pls.
Thanks

Area of ABCD = 959.93 units²

Step-by-step explanation:

a). By applying Sine rule in the ΔABD,

 Sin∠DBA =

 m∠DBA =

 m∠DBA = 45.64°

 Therefore, m∠ADB = 180° - (110° + 45.64°) = 24.36°

                  m∠ADB = 24.36°

c). Area of ABCD = Area of ΔABD + Area of ΔBCD

  Area of ΔABD = AD×BD×Sin()

                           = 35×46Sin(12.18)

                           = 339.68 units²

   Area of ΔBCD = BD×BC×Sin()°

                            = 46×27×(0.4994)

                            = 620.25 units²

   Area of ABCD = 339.68 + 620.25

                            = 959.93 units²

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}

a

step-by-step explanation: