Given :-
- The general term of a sequence is given by aₙ=43-3(n-1) .
To Find :-
- The first four terms of the sequence.
Solution :-
The given expression is 
→ aₙ=43-3(n-1)
where n > 0
<u>Finding</u><u> the</u><u> </u><u>first </u><u>term </u><u>:</u>
Substituting n = 1 , we have ,
→ T1 = 43 - 3(1-1)
→ T1 = 43 - 3*0
→ T1 = 43 - 0 = 43
<u>Finding</u><u> the</u><u> </u><u>second</u><u> </u><u>term </u><u>:</u>
Substituting n = 2 , we have,
→ T2 = 43 -3(2-1)
→ T2 = 43 -3*1
→ T2 = 43 -3 = 40
<u>Finding</u><u> </u><u>the </u><u>third </u><u>term</u><u> </u><u>:</u>
Substituting n = 3 , we have,
→ T3 = 43 -3(3-1)
→ T3 = 43 -3*2
→ T3 = 43 -6 = 37
<u>Finding</u><u> the</u><u> </u><u>fourth</u><u> </u><u>term </u><u>:</u>
→ T4 = 43 -3(4-1)
→ T4 = 43 -3*3
→ T4 = 43-9 = 34
<u>Hence</u><u> the</u><u> </u><u>first</u><u> </u><u>four</u><u> terms</u><u> of</u><u> </u><u>the</u><u> </u><u>sequence</u><u> </u><u>are </u><u>4</u><u>3</u><u> </u><u>,</u><u> </u><u>4</u><u>0</u><u> </u><u>,</u><u> </u><u>37</u><u> </u><u>and </u><u>34</u><u> </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em><em> </em><em>Let </em><em>me</em><em> know</em><em> if</em><em> you</em><em> </em><em>need </em><em>further</em><em> </em><em>clarification</em><em> </em><em>.</em>
C is correct bc if you add 10 to x you get the value of y
Answer:
(4 x 3) x 2=
=12 x 2
=24
Step-by-step explanation:
Hey there! I'm happy to help!
Let's call our fraction x.
x=3.777......
We want to get rid of the repeating stuff. If we multiply our fraction by 10, we get 10x=37.7777.
We see that if we subtract x from 10x, the repeating numbers will be cancelled out. Let's do this.
10x=37.7777
-
x=3.77777
=
9x=34
We divide both sides by 9.
x=34/9=3 7/9.
Have a wonderful day! :D