Evaluate expression for t=0: The series is a finite arithmetic series with two terms, t=0 and t=1. We will evaluate the expression (4-t) for each value of t and then sum the results. Evaluate expression for t=1: First, we substitute t=0 into the expression (4-t) to get the first term of the series. This gives us 4-0, which equals 4. Sum the terms of the series: Next, we substitute t=1 into the expression (4-t) to get the second term of the series. This gives us 4-1, which equals 3. Now, we add the two terms of the series together. The sum is 4 (from t=0) plus 3 (from t=1), which equals 7.

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$\sum_{t=0}^{1}(4-t)=$

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Calculus

Find derivatives using the chain rule I

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\sum_{t=0}^{1}(4-t)=

Evaluate expression for $t=0$ : The series is a finite arithmetic series with two terms, $t=0$ and $t=1$ . We will evaluate the expression $(4-t)$ for each value of $t$ and then sum the results.
Evaluate expression for $t=1$ : First, we substitute $t=0$ into the expression $(4-t)$ to get the first term of the series. This gives us $4-0$ , which equals $4$ .
Sum the terms of the series: Next, we substitute $t=1$ into the expression $(4-t)$ to get the second term of the series. This gives us $4-1$ , which equals $3$ .
Sum the terms of the series: Next, we substitute $t=1$ into the expression $(4-t)$ to get the second term of the series. This gives us $4-1$ , which equals $3$ . Now, we add the two terms of the series together. The sum is $4$ (from $t=0$ ) plus $3$ (from $t=1$ ), which equals $7$ .