An= mth term.
an=a₁+(n-1)*d
a₁₂=41
a₁₅=140
a₁₂=41
41=a₁+(12-1)*d
41=a₁+11d
a₁+11d=41 (1)
a₁₅=140
140=a₁+(15-1)*d
140=a₁+14d
a₁+14d=140 (2)
With the equiations (1) and (2) build a system of equations
a₁+11d=41
a₁+14d=140
we solve it.
-(a₁+11d=41)
a₁+14d=140
--------------------
3d=99 ⇒d=99/3=33
a₁+11d=41
a₁+(11*33)=41
a₁+363=41
a₁=41-363=-322
an=a₁+(n-1)*d
an=-322+(n-1)*33
an=-322+33n-33
an=-355+33n
an=-355+33n
To check:
a₁₂=-355+33*12=-355+396=41
a₁₅=-355+33*15=-355+495=140.
Answer:
x=1/8y+9/8
y=<u>y=−8x+9</u>
Step-by-step explanation:
<u>Let's solve for y.</u>
<u>8x+y=9</u>
<u>Step 1: Add -8x to both sides.</u>
<u>8x+y+−8x=9+−8x</u>
<u>y=−8x+9</u>
<u />
<u>Let's solve for x.</u>
<u>8x+y=9</u>
<u>Step 1: Add -y to both sides.</u>
<u>8x+y+−y=9+−y</u>
<u>8x=−y+9</u>
<u>Step 2: Divide both sides by 8.</u>
<u>8x/8=−y+9/8</u>
<u>x=−1/8y+9/8</u>
<u />
<u />
The parents could have 3 boys and 3 girls in 400 ways
<h3>How to determine the number of ways?</h3>
The given parameters are
Children, n = 6
Boys, r = 3
Girls, r =3
The number of combination is:

So, we have:

Apply the combination formula

This gives
Ways = 400
Hence, the parents could have 3 boys and 3 girls in 400 ways
Read more about combination at:
brainly.com/question/11732255
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