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What is the dimension of 2 X 2R?

Updated: 12/18/2022
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The same as the dimension of R

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Q: What is the dimension of 2 X 2R?
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What is the dimension of 2R photo?

2R = 2.5 inch x 3.5 inch


What is the answer for 2r plus 5 -1?

2r + 4 or 2*(r+2)


2n plus 3 is an arithmetic sequence how many terms should be taken starting from the first term so that the ratio of the sum of the first third of these terms to the rest of the terms is 7 to 44?

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.


What is 3r-8 equals2-2r?

If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2


What is C equals 2r for r?

If: C = 2r Then: r = C/2

Related questions

What is the dimension of 2R photo?

2R = 2.5 inch x 3.5 inch


What is the radius of a full adult size NBA basketball ball?

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.


What are the factors of the monomial 2r³s?

2 x r x r x r x s = 2r3s


What is the answer for 2r plus 5 -1?

2r + 4 or 2*(r+2)


What is the dimension of an 8 cubic box?

If it's a cube, the dimensions are 2 x 2 x 2.


Identify the boundaries of the solution of this inequality 2r 2?

2r


What is 4r2-25 factored?

4r2-25 = (2r)2-52 = (2r-5)(2r+5)


If you double the radius of a circle what happens to the area?

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.


2n plus 3 is an arithmetic sequence how many terms should be taken starting from the first term so that the ratio of the sum of the first third of these terms to the rest of the terms is 7 to 44?

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.


How do you do absolute 2r plus 9 absolute equals 5?

|2r + 9| = 5 So 2r + 9 = 5 or 2r + 9 = -5 so that 2r = -4 or 2r = -14 r = -2 or r = -7


How do you solve 2r-7 equals 9?

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


What is augmented vector?

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.