Advanced Math/math q
Expert: Sherman D. - 1/5/2008
QuestionThe sides of the large square are 16 cm. The midpoints of the sides are joined to form a new square. Find the sum of the areas of all the squares. There are 7 squares
The interior angles of a hexagon are in an arithmetic sequence. The largest angle is 130 degrees. What are the other angles.
THe first the terms of an arithmetic series have a sum of 24 and a product of 312. What is the fourth term of the series?
These questions have really got me. Fishised every other question but I don't even know where to start with these 3.
AnswerThe sides of the large square are 16 cm. The midpoints of the sides are joined to form a new square. Find the sum of the areas of all the squares. There are 7 squares
On this one, can you describe it more. are you saying that you have 7 squares all of which are formed by the connection of the midpoints of 2 sides, so you have "A" square inside "A" square, in sides "A" square, etc... forming one large square(the original square), and inside that is another square, and inside that square is another square.
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The interior angles of a hexagon are in an arithmetic sequence. The largest angle is 130 degrees. What are the other angles.
The interior angles of a hexagon add up to be 720, so this gives us
S(n) = (n/2)(a1 + an)
S(6) = (6/2)(130 + a(6))
720 = 3(130 + an)
720 = 390 + 3a(6)
3a(6) = 330
a(6) = 110
so now this gives us
an = a1 - d(n - 1)
110 = 130 - d(6 - 1)
-20 = -5d
d = 4
so now this us
a(n) = 130 - 4(n - 1)
so the numbers are
a(1) = 130
a(2) = 130 - 4(2 - 1) = 130 - 4(1) = 130 - 4 = 126
a(3) = 130 - 4(3 - 1) = 130 - 4(2) = 130 - 8 = 122
a(4) = 130 - 4(4 - 1) = 130 - 4(3) = 130 - 12 = 118
a(5) = 130 - 4(5 - 1) = 130 - 4(4) = 130 - 16 = 114
a(6) = 130 - 4(6 - 1) = 130 - 4(5) = 130 - 20 = 110
ANS : 110, 114, 118, 122, 126, and 130
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you say "the first the terms" do you mean "the first three terms".
either way
24 = (n/2)(a1 + a(n))
a(n) * a(n + 1) * a(n + 2) * a(n + 3) = 312
I'm going to need some more info if you can profide it.