In an arithmetic sequence:
Tn=t₁+(n-1)d
t₄=t₁+(4-1)d=t₁+3d
t₅=t₁+(5-1)d=t₁+4d
t₆=t₁+(6-1)d=t₁+5d
t₄+t₅+t₆=(t₁+3d) +(t₁+4d)+(t₁+5d)=3t₁+12d
Therefore:
3t₁+12d=300 (1)
t₁₅=t₁+(15-1)d=t₁+14d
t₁₆=t₁+(16-1)d=t₁+15d
t₁₇=t₁+(17-1)d=t₁+16d
t₁₅+t₁₆+t₁₇=(t₁+14d)+(t₁+15d)+(t₁+16d)=3t₁+45d
Therefore:
3t₁+45d=201 (2)
With the equations (1) and (2) we make an system of equations:
3t₁+12d=300
3t₁+45d=201
we can solve this system of equations by reduction method.
3t₁+12d=300
-(3t₁+45d=201)
-----------------------------
-33d=99 ⇒d=99/-33=-3
3t₁+12d=300
3t₁+12(-3)=300
3t₁-36=300
3t₁=300+36
3t₁=336
t₁=336/3
t₁=112
Threfore:
Tn=112+(n-1)(-3)
Tn=112-3n+3
Tn=115-3n
Now, we calculate T₁₈:
T₁₈=115-3(18)=115-54=61
Answer: T₁₈=61
Weird way to write it but alright! (Sideways)
19pq^-2 x 5pq^6 = ?
These problems are pretty much single operations between each of the variables / constants.
So it's like this:
(19*5)(p*p)(q^-2*q^6) = ?
19*5 is 95.
For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2.
p*p is p^2
For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this:
-2 + 6 = 4
(The two is negative, because the power is negative 2)
So, q^4.
Our final answer is all of the combined.... like a so:
95p^2q^4
Yes, Brainly is a very useful app. I love helping people, and it allows me to gain some help from others when needed :)
Answer:
- The greatest amount of rainfall is represented by the greatest number
- The least amount of rainfall is represented by the least number
Step-by-step explanation:
