Answer:
a
Step-by-step explanation:
Im just going out on a wing here but is one options 40 + 45
<span>Step 1. Let n be an integer.
Step 2. Let n+1 and n+2 be the next two consecutive integers.
Step 3. Let n+n+1=2n+1 be the sum of the two smallest integers.
Step 4. Let 2(2n+1) be twice the sum of the two smallest integers.
Step 5. Let 3(n+2) be three times the largest number
Step 6. Then using Steps 3 and 4, 2(2n+1)=3(n+2)+10 since twice the
sum of the two smallest numbers is 10 more than three times the largest
number.
Step 7. Solving yields the following steps:
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