To solve the given expression 3t2+5t+63t^2 + 5t + 63t2+5t+6 for t=−3t = -3t=−3, we need to substitute −3-3−3 for ttt in the expression and then evaluate it.
We start with the expression 3t2+5t+6 3t^2 + 5t + 6 3t2+5t+6. To find its value when t=−3 t = -3 t=−3, we substitute −3-3−3 for t t t:
3(−3)2+5(−3)+6 3(-3)^2 + 5(-3) + 6 3(−3)2+5(−3)+6
First, calculate (−3)2 (-3)^2 (−3)2:
(−3)2=9 (-3)^2 = 9 (−3)2=9
Next, multiply by 3:
3×9=27 3 \times 9 = 27 3×9=27
Then, calculate 5(−3) 5(-3) 5(−3):
5×(−3)=−15 5 \times (-3) = -15 5×(−3)=−15
Finally, add all the terms together:
27+(−15)+6=27−15+6=18 27 + (-15) + 6 = 27 - 15 + 6 = 18 27+(−15)+6=27−15+6=18
The value of the expression 3t2+5t+6 3t^2 + 5t + 6 3t2+5t+6 when t=−3 t = -3 t=−3 is 18\boxed{18}18.
The answer is (d) 18.
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