<span>So you want to factor this polynom thus it will look like something like this
(x+a)(x+b) where a and b can be positive or negative
now in our question the polynom ends with +35 that has no x in it. This part can only come from a*b. Two numbers multiplied can only be positive if they are both positive or if they are both negative.
</span><span>-26x this part comes from ax+bx part of the above equation. If both a and b are poitive than the result needs to be positive as well, this combining the two arguements we come to the conclusion that they are both negative.
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A student can only belong to one age group! You can't be 5 and 11 :p
Answer for problem 46 is choice A
Answer for problem 47 is choice B
Answer for problem 48 is choice E
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Work Shown
Problem 46
Equation 1: 3x+y = 17
Equation 2: x+3y = -1
Add equation 1 to equation 2 to get 4x+4y = 16. Divide every term by 4 to get x+y = 4. Then finally multiply both sides by 3 to get 3x+3y = 12
That shows why the answer is choice A
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Problem 47)
If y hours pass by, then y-(2/3)y=y/3 is the time value (2/3)y hours ago
So,
Distance = rate*time
d = r*t
d = x*(y/3)
d = (xy)/3
That's why the answer is choice B
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Problem 48)
Let L1,L2,L3 be the three lists where
L1 = {a1,a2,a3,...,ak} there are k scores here
L2 = {a1,a2,...,a10} there are 10 scores here
L3 = {a11,a12,...,ak} the remaining k-10 scores
S(L1) = sum of the scores in list L1
M(L1) = mean of L1 = 20 = S(L1)/k
M(L2) = mean of L2 = 15 = S(L2)/10
S(L1) = 20k
S(L2) = 150
S(L1) = S(L2)+S(L3)
M(L1) = [S(L2)+S(L3)]/k
20 = [150+S(L3)]/k
20k = 150+S(L3)
S(L3) = 20k-150
M(L3) = [S(L3)]/(k-10)
M(L3) = (20k-150)/(k-10)
So that shows why the answer is choice E