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question:Solve the system of linear equations given below: begin{align*} 7x - 22y + 23 &= 0 19x - 14z + 22 &= 0 -22x + 15y - 12z + 10 &= 0 end{align*}

answer:The solution to the system of equations is: begin{align*} x &= frac{1051}{5161} y &= frac{5730}{5161} z &= frac{19073}{10322} end{align*}

question:What is the total number of hydrogen atoms in two molecules of the compound #Ca(OH)_4#?

answer:Each molecule has 2 hydrogen atoms in the brackets, and the subscript 4 multiplies that by 4, giving 8 hydrogen atoms per molecule. Since there are two molecules, we multiply by 2 to get #8xx2=16#. Therefore, there are 16 hydrogen atoms.

question:Compute the binomial coefficient binom{10997}{10996}.

answer:The binomial coefficient binom{n}{k} represents the number of ways to choose k elements from a set of n elements, without regard to order. In this case, we have n = 10997 and k = 10996. Using the formula for the binomial coefficient, we have: binom{10997}{10996} = frac{10997!}{10996! cdot 1!} = frac{10997 cdot 10996!}{10996! cdot 1} = 10997 However, since 10996 is one less than 10997, we can simplify further: binom{10997}{10996} = 10997 - 10996 = 1 Therefore, the binomial coefficient binom{10997}{10996} is equal to 1. The answer is 1

question:How many permutations a_1,a_2,a_3,..., a_{100} of the integers from 1 to 100 are there such that none of the sums a_1,a_1+a_2,a_1+a_2+a_3,dots ,a_1+a_2+dots + a_{100} are divisible by 3?

answer:Hint: For a moment forget all the multiples of 3 that appear in the sequence. This leaves 34 numbers that leave a reminder of 1 and 33 numbers that leave a reminder of 2. So this problem is all about finding the number of sequences containing 34 ones and 33 twos such that the sum of the first k numbers is not divisible by 3. (After finding such a sequence you need to insert the multiples of 3 into the sequence. This will produce all the sequences that are sought for.) The sequence either begins with 2 or 1. If it begins with 1 then the sequence will have to proceed as follows, 1,1,2,1,2,1,2,1,2,1,2 dots That should give you a start.