(10) \( \sum_{t=-2}^{30}(4 t+6) \quad \) (11) \( \sum_{p=0} 3^{4-p} \)

To tackle the first summation, \( \sum_{t=-2}^{30}(4t+6) \), we can simplify it using the formula for the sum of an arithmetic series. The expression \( 4t + 6 \) creates a linear series of terms evaluated at \( t = -2, -1, ..., 30 \). The first term when \( t = -2 \) is \( 4(-2) + 6 = -8 + 6 = -2 \), and the last term when \( t = 30 \) is \( 4(30) + 6 = 120 + 6 = 126 \). The number of terms can be calculated as \( 30 - (-2) + 1 = 33 \). Now, the average of the first and last term is \( \frac{-2 + 126}{2} = 62 \). Thus, the entire sum becomes \( 33 \times 62 = 2046 \). For the second summation, \( \sum_{p=0}^{4} 3^{4-p} \), we note this is a series of exponents as \( 3^4, 3^3, 3^2, 3^1, 3^0 \) as \( p \) takes values from 0 to 4. This simplifies to \( 81 + 27 + 9 + 3 + 1 = 121 \). Alternatively, recognizing that this series is a geometric series with ratio \( r = \frac{1}{3} \) and \( n = 5 \), we can apply the geometric series formula \( S_n = a \frac{1 - r^n}{1 - r} \) where \( a = 81 \). Hence, the sum also yields \( 121 \). Final Results: (10) \( \sum_{t=-2}^{30} (4t + 6) = 2046 \) (11) \( \sum_{p=0}^{4} 3^{4-p} = 121 \)