Hank started a savings account in june of 2002. on june 2006, he had $3,800. on june 2014, he had $8,600. if hank's saving is modeled by a linear function, what was his initial deposit?

$1,400

Step-by-step explanation:

slope =  

8,600 - 3,800

2014 - 2006

= 600

y = mx + b

3,800 = 600(2006) + b

b = -1,199,800

thus,

y = mx + b

y = 600x - 1,199,800

y = 600(2002) - 1,199,800

y = 1,400

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92