Answer:
57.09
Step-by-step explanation:
Hello :
(-6a)-(-3a) =(-3a)-(0) = (0)-(3a) = (3a)-(6a) = -3a (arthmetic<span> sequence, the common - diff is : d= -3a the first term is : U1 = 6a
</span><span>a general rule for the n th term is : Un =U1+(n-1)d
Un = 6a +(n-1)(-3a) =
Un = 6a -3an+3a
</span>Un = -3an +9a
Answer:
c is the correct awnser triangle ghj coordinates
<span>2 13/20, that should work</span>
To find how many Japanese yen $1 was worth in 2010 and 2013, you will create ratios with the information we know and then solve for the number of yen.
2010:
100 yen x yen (use cross products to solve for x yen)
$0.88 = $1
100 = 0.88x
0.88 0.88
x = 113.6 yen
In 2010, $1 was worth approximately 114 yen.
2013:
100 yen x yen
$0.93 = $1
100 = 0.93x
0.93 0.93
x = 107.5
In 2013, $1 was worth approximately 108 yen.