answer:(I) For the arithmetic sequence {a_{n}}, we know that a_{1}=3 and a_{2}+a_{5}=11. Since a_{2}+a_{5}=a_{1}+a_{6}, we can deduce that a_{6}=8. Therefore, the common difference between terms is 1, and the general term formula for the sequence is a_{n}=n+2. (II) Using the general term formula a_{n}=n+2, we can rewrite c_{n} as c_{n}=2^{a_{n}-2}+n=2^{n+2-2}+n=2^{n}+n. To find the sum of the first 10 terms for the sequence {c_{n}}, we compute: S_{10}=sum_{n=1}^{10}(2^{n}+n)=sum_{n=1}^{10}2^{n}+sum_{n=1}^{10}n=(2+2^{2}+2^{3}+dots+2^{10})+(1+2+dots+10). The first sum can be evaluated using the formula for the sum of a geometric series: sum_{n=1}^{10}2^{n}=2frac{1-2^{10}}{1-2}=2(2^{10}-1). The second sum is the sum of an arithmetic series, which can be computed as: sum_{n=1}^{10}n=frac{10times11}{2}=55. Therefore, S_{10}=2(2^{10}-1)+55=boxed{2^{11}+53}.

answer:Let's calculate Donna's earnings from dog walking, babysitting, and then we can find out how much she made at the card shop. Dog walking: She worked 2 hours every morning for 10.00 an hour, 5 days a week. Earnings from dog walking = 2 hours/day * 10.00/hour * 5 days/week = 100.00/week Babysitting: She made 10.00 an hour and worked 4 hours every Saturday. Earnings from babysitting = 4 hours * 10.00/hour = 40.00/week Now, let's subtract the earnings from dog walking and babysitting from the total earnings to find out how much she made at the card shop. Total earnings = 305.00 Earnings from dog walking and babysitting = 100.00 + 40.00 = 140.00 Earnings from the card shop = Total earnings - Earnings from dog walking and babysitting Earnings from the card shop = 305.00 - 140.00 = 165.00 Now, we need to find out how many hours she worked at the card shop and how much she made per hour. She worked there 5 days a week for 2 hours each day. Total hours at the card shop = 5 days/week * 2 hours/day = 10 hours/week Now we can calculate her hourly wage at the card shop. Hourly wage at the card shop = Earnings from the card shop / Total hours at the card shop Hourly wage at the card shop = 165.00 / 10 hours = 16.50/hour Donna made boxed{16.50} per hour working at the card shop.

answer:1. **Dot product condition**: The equation from the dot product is: [ 3x + 2y + 4z = 20. ] 2. **Cross product condition**: The cross product of vectors mathbf{a} and mathbf{b} is computed as: [ mathbf{a} times mathbf{b} = begin{pmatrix} 3 2 4 end{pmatrix} times begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} 2z - 4y 4x - 3z 3y - 2x end{pmatrix}. ] Setting this equal to the given vector begin{pmatrix} 1 -15 5 end{pmatrix} results in a system: [ begin{align*} 2z - 4y &= 1, 4x - 3z &= -15, 3y - 2x &= 5. end{align*} ] 3. **Solve the system of equations**: [ begin{align*} 2z - 4y &= 1, quad text{(I)} 4x - 3z &= -15, quad text{(II)} 3y - 2x &= 5, quad text{(III)} 3x + 2y + 4z &= 20, quad text{(IV)} end{align*} ] Solve equations (I) and (III) together to express z and x in terms of y, and then substitute into (IV) to find y. Subsequently find z from (I) and x from (III). After solving these, we find: [ begin{align*} x &= 7, y &= -2, z &= 3. end{align*} ] Hence, mathbf{b} = boxed{begin{pmatrix} 7 -2 3 end{pmatrix}}.

answer:(1) The negations of the propositions p and q are: - neg p: There exists an x_0 in mathbb{R} such that x_0^2 + x_0 leq a. - neg q: For all x in mathbb{R}, x^2 + 2ax + 2 - a neq 0. (2) For proposition p: Since x^2 + x = (x + frac{1}{2})^2 - frac{1}{4} which is greater than or equal to -frac{1}{4} for any real number x, it follows that a < -frac{1}{4}. For proposition q: There exists an x in mathbb{R} that satisfies x^2 + 2ax + 2 - a = 0. Therefore, the discriminant must be non-negative: Delta = 4a^2 - 4(2-a) geq 0. Solving this yields a geq 1 or a leq -2. Therefore, for the negations we have: - neg p: a geq -frac{1}{4}. - neg q: -2 < a < 1. Since the proposition (neg p) vee q is false, both neg p and q must be false. Therefore, both p and neg q must be true. Hence, we have the system: [ left{ begin{aligned} &a < -frac{1}{4} &-2 < a < 1 end{aligned} right. ] Solving this system gives: [ -2 < a < -frac{1}{4} ] Therefore, the range of values for the real number a is boxed{-2 < a < -frac{1}{4}}.