8 bra cause that’s the answer I got for sure
Let 1993 = time 0 = 0.
Let 1999 = time 6 = 6
Let 2012 = time 19 = 19
So, a = 171 (million). First solve for k.
176 = 171 e^k6
176/171 = e^(k*6)
ln (176/171) = 6k
k = 1/6 ln (176/171)
So, in 2012 we have: P(19) = 171 e^(19k), where k = 1/6 ln (176/171)
Hope this helped!
<u>Answer:</u> 5 years!
<u>Reasoning:</u>
Year 1: 2700 x .70=1890
Year 2: 1890 x .70= 1323
Year 3: 1323 x .70= 926.1
Year 4: 926.1 x .70= 648.27
Year 5: 648.27 x .70=453.79
Answer:
The nth term of an AP will be 27 -7n.
Step-by-step explanation:
First five terms of the Arthemetic Sequence is given to us , which is 26 , 19 , 12 , 5
Hence here Common Difference can be found by subtracting two consecutive terms . Here which is 19 - 26 = (-7) .
Here first term is 26 .
And the nth term of an AP is given by ,
★ T_n = a + ( n - 1) d
<u>Subst</u><u>ituting</u><u> respective</u><u> values</u><u> </u><u>,</u>
⇒ T_n = a + ( n - 1 )d
⇒ T_n = 26 + (n - 1)(-7)
⇒ T_n = 26 -7n+1
⇒ T_n = 27 - 7n
<h3>
<u>Hence </u><u>the</u><u> </u><u>nth</u><u> </u><u>term</u><u> of</u><u> an</u><u> </u><u>AP</u><u> </u><u>can</u><u> </u><u>be</u><u> </u><u>found </u><u>using </u><u>T_</u><u>n</u><u> </u><u>=</u><u> </u><u>2</u><u>7</u><u> </u><u>-</u><u> </u><u>7</u><u>n</u><u>. </u></h3>
I believe it's b , hopefully this helps
