Answer:
7. 10 8. 20 9. 100 thats all i know
Step-by-step explanation:
Answer:
To prove that 3·4ⁿ + 51 is divisible by 3 and 9, we have;
3·4ⁿ is divisible by 3 and 51 is divisible by 3
Where we have;
= 3·4ⁿ + 51
= 3·4ⁿ⁺¹ + 51
- = 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ
- = 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ
∴ - is divisible by 9
Given that we have for S₀ = 3×4⁰ + 51 = 63 = 9×7
∴ S₀ is divisible by 9
Since - is divisible by 9, we have;
- 
= 
- is divisible by 9
Therefore is divisible by 9 and is divisible by 9 for all positive integers n
Step-by-step explanation:
It should give you the option to do so.
Suppose the number is y >>>>>So y/2 +2y/3 <14
multiply both sides by 6>>>>>>>>>. 6y/2 +6x2y/3 <14x6
3y + 4y < 84
7y <84
y <84/7
y < 12
Change you’ll receive is = 0.09
you can only buy 11 because if you divide 20/11 it will give you 1.81 and if you multiply 11x 1.81 it will give you 19.91 which is close to $20 & that’s the most you can buy .
