Answer:
- an = 2·(4/5)^(n-1)
- 2, 8/5, 32/25, 128/125
Step-by-step explanation:
The sum of an infinite geometric series is ...
S = a1/(1 -r)
where r is the common ratio. The sum will only exist if |r| < 1.
The problem statement tells us S = 10 and a1 = 2, so we have ...
10 = 2/(1 -r)
r = 1 -2/10 = 4/5
So the n-th term of the series is ...
an = a1·r^(n-1)
an = 2·(4/5)^(n-1)
For values of n = 1 to 4, the terms are ...
2, 8/5, 32/25, 128/125
10 because first brackets is 0 second brackets is 10 15*0 is 0 0+10 is 10
Answer: 
Move all terms containing y to the left, all other terms to the right
<u>Add -14y to each side of the equation</u>
<u></u>
<u></u>
<u></u>
<u>Combine like terms: 5y + -14y = -9y
</u>
<u></u>
<u></u>
<u></u>
<u>Combine like terms: 14y + -14y = 0
</u>
<u></u>
<u></u>
<u></u>
<u>Add '20' to each side of the equation</u>
<u></u>
<u></u>
<u></u>
<u>Combine like terms: -20 + 20 = 0
</u>
<u></u>
<u></u>
<u></u>
<u>Combine like terms: 7 + 20 = 27
</u>
<u></u>
<u></u>
<u></u>
<u>Divide each side by -9</u>
<u></u>
<u></u>
How many triangles can<span> you </span>construct<span> given three </span>angle<span> measures whose sum is 180°? The sum of the</span>angle<span> measures of any </span>triangle<span> is 180°. You </span>can<span> use a protractor to </span>construct<span> a </span>triangle<span> given three</span>angle<span> measures. </span>90 90<span>. </span>80<span> 100. 70 110. 60. 120. 50. 130. 40. 140. 30. 150. 20. 160. </span>10. 170. 0. 180. 100.80<span>. 110 70. 12.</span>
