2R = 2.5 inch x 3.5 inch
2r + 4 or 2*(r+2)
The sum of an arithmetic series of n terms is: S(n) = n(2a1 + (n - 1)d) / 2 where a1 is the first term and d is the difference between terms. The key to solving this is to realise that two sums are required, S1 and S2 such that: S1 = S(r) S2 = S(3r) - S(r). But that this second sum can also be stated as: S2 = sum of 2r terms starting at term ar+1 So that the problem then becomes: Find two sums S1 and S2 of the sequence an = 2n + 3 such that the first (S1) has r terms starting at a1 and the second (S2) has 2r terms starting at ar+1, with the ratio of S1 : S2 = 7 : 44. This gives: S1 = r (2a1 + (r - 1)d) / 2 = r(2(2x1 + 3) + (r - 1)2) / 2 = r(2r + 8) / 2 = r(r + 4) S2 = 2r(2ar+1 + (2r - 1)d) = 2r(2(2(r + 1) + 3) + (2r - 1)2) / 2 = 2r(2(2r + 5) + (2r - 1)2) / 2 = 2r(2r + 5 +2r - 1) = 2r(4r + 4) = 8r(r + 1) As S1 : S2 = 7 : 44, 44 S1 = 7 S2 => 44 r(r + 4) = 7 x 8r(r + 1))0 => 11(r + 4) = 14(r + 1) => 11r + 44 = 14r + 14 => 3r = 30 => r = 10 As r is the number of terms in S1 it is a third of the number of terms required, so 3 x 10 = 30 terms are needed. To check: a1 = 2 x 1 + 3 = 5 S(30) = 30(2 x 5 +(30-1) x 2) / 2 = 30 x 34 = 1020 S(10) = 10(2 x 5 + (10-1) x 2) / 2 = 10 x 14 = 140 Sum of 1st third S1 = S(10) = 140, sum of rest S2 = S(30) - S(10) = 1020 - 140 = 880 Ratio of S1 : S2 = 140 : 880 = 14 : 88 = 7 : 44 Note that the second sum S2 is also 20 terms starting at a11 = 2 x 11 + 3 = 25: S2 = 20(2 x 25 + 19 x 2)/2 = 20 x 44 = 880.
If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2
If: C = 2r Then: r = C/2
2R = 2.5 inch x 3.5 inch
Diameter = 2 x Radius Circumference = π x Diameter= π x (2 x Radius) 28.5 = π x 2R 28.5 / π = 2R 28.5/3.14159 = 2R = 9.0718 OR 9.1 Radius = 9.1/2 = 4.5 inches.
2 x r x r x r x s = 2r3s
2r + 4 or 2*(r+2)
If it's a cube, the dimensions are 2 x 2 x 2.
2r
4r2-25 = (2r)2-52 = (2r-5)(2r+5)
If you double the original radius, the area will become 4 times what it was. The area of a circle is pi*r^2 = A. If our new radius (lets say x) is double (x=2r), we have pi*x^2 = pi*(2r)^2, which equals 4*pi*r^2 = 4A, or 4 times the original area.
The sum of an arithmetic series of n terms is: S(n) = n(2a1 + (n - 1)d) / 2 where a1 is the first term and d is the difference between terms. The key to solving this is to realise that two sums are required, S1 and S2 such that: S1 = S(r) S2 = S(3r) - S(r). But that this second sum can also be stated as: S2 = sum of 2r terms starting at term ar+1 So that the problem then becomes: Find two sums S1 and S2 of the sequence an = 2n + 3 such that the first (S1) has r terms starting at a1 and the second (S2) has 2r terms starting at ar+1, with the ratio of S1 : S2 = 7 : 44. This gives: S1 = r (2a1 + (r - 1)d) / 2 = r(2(2x1 + 3) + (r - 1)2) / 2 = r(2r + 8) / 2 = r(r + 4) S2 = 2r(2ar+1 + (2r - 1)d) = 2r(2(2(r + 1) + 3) + (2r - 1)2) / 2 = 2r(2(2r + 5) + (2r - 1)2) / 2 = 2r(2r + 5 +2r - 1) = 2r(4r + 4) = 8r(r + 1) As S1 : S2 = 7 : 44, 44 S1 = 7 S2 => 44 r(r + 4) = 7 x 8r(r + 1))0 => 11(r + 4) = 14(r + 1) => 11r + 44 = 14r + 14 => 3r = 30 => r = 10 As r is the number of terms in S1 it is a third of the number of terms required, so 3 x 10 = 30 terms are needed. To check: a1 = 2 x 1 + 3 = 5 S(30) = 30(2 x 5 +(30-1) x 2) / 2 = 30 x 34 = 1020 S(10) = 10(2 x 5 + (10-1) x 2) / 2 = 10 x 14 = 140 Sum of 1st third S1 = S(10) = 140, sum of rest S2 = S(30) - S(10) = 1020 - 140 = 880 Ratio of S1 : S2 = 140 : 880 = 14 : 88 = 7 : 44 Note that the second sum S2 is also 20 terms starting at a11 = 2 x 11 + 3 = 25: S2 = 20(2 x 25 + 19 x 2)/2 = 20 x 44 = 880.
2r - 7 = 9 (+7) (+7) >> add 7 to both sides so that only 2r is on that side 2r = 16 2r/2 = 16/2 >> divide both sides by 2 to isolate the variable r=8 >>16 divided by 2 is 8 which is your answer
|2r + 9| = 5 So 2r + 9 = 5 or 2r + 9 = -5 so that 2r = -4 or 2r = -14 r = -2 or r = -7
An augmented vector is a vector that is augmented with an extra dimension. This new dimension always takes on the value of 1. e.g. X = (5, 2) X' = (5, 2, 1) where X' is the augmented form of vector X.