( 11s + 3 )
click on the clocks that are at the times 2 3 5 7 11
B = bigger numberS = smaller numberYou told us:B = 3SB - S = 2,184Substitute 3S in place of B in the second statement:3S - S = 2,1842S = 2,184S = 1,092Their sum is B+S, and we know that B=3S.B + S = 3S + S = 4S = 4,368.
Well 3 20's, 3 10's and 3 5's would be $105.
25+11=36: Let f and s represent the first and second numbers respectively. The statement of the problem yields two equations: f + s =36 and f = 3 + 2s. Substituting the function given in the second equation for f into the first equation yields 3 + 2s + s = 36, or (subtracting 3 from each side and merging the s terms, 3s = 33 or s = 11. Then f + 11 = 36 (substituting the value for s into the first equation), or f = 25.
The sum of all the angles found on any polygon is always found by this forumla: s = (n-2)180 "n" is the number of angles on the polygon; "s" is the sum. Therefore, replace "n" with "5": s = (5-2)180 s = 3 x 180 s = 540 The sum of the angles of a pentagon (5-sided polygon) is 540 degrees.
11s + 3
To find the number of sides ( n ) in a polygon given the sum of its interior angles, you can use the formula ( S = (n - 2) \times 180 ), where ( S ) is the sum of the interior angles. Setting ( S = 1620 ), we get: [ 1620 = (n - 2) \times 180 ] Solving for ( n ): [ n - 2 = \frac{1620}{180} = 9 \ n = 11 ] Thus, the polygon has 11 sides.
=7(x+s)
11 (interior angles total 18 x 90)
click on the clocks that are at the times 2 3 5 7 11
The sum of the interior angle measures of a polygon can be calculated using the formula ( S = (n - 2) \times 180 ), where ( n ) is the number of sides. To find the number of sides for which the sum of the interior angles is 1620 degrees, we can set up the equation: ( (n - 2) \times 180 = 1620 ). Solving for ( n ), we get ( n - 2 = 9 ), so ( n = 11 ). Therefore, a polygon with a sum of interior angles measuring 1620 degrees has 11 sides, making it an 11-sided polygon, or hendecagon.
The sum of all odd numbers from 1 to 59 can be calculated using the formula for the sum of an arithmetic series. The series consists of the first 30 odd numbers (1, 3, 5, ..., 59). The sum can be calculated as ( S = n/2 \times (first\ term + last\ term) ), where ( n ) is the number of terms. Thus, the sum is ( 30/2 \times (1 + 59) = 15 \times 60 = 900 ).
The sum is 50 and 31/40 The difference is 1/40
It is: 1*11 = 11 which is a prime number
111 + 11 + 1 = 123
B = bigger numberS = smaller numberYou told us:B = 3SB - S = 2,184Substitute 3S in place of B in the second statement:3S - S = 2,1842S = 2,184S = 1,092Their sum is B+S, and we know that B=3S.B + S = 3S + S = 4S = 4,368.
Let the inegers be x+1, x+3, and x+5, these are 3 consecutive odds, we could have used x and x+1 and x+3, if x was odd Now the next part is often confusing. So let's say the sum was S, we have S=2(x+1)-5. Now, S= 3x+9 So we have 3x+9=2x+2-5 or x=-9-5+2 so x=-12 x+1=-11 which is odd x+3=-9 x+5=-7 The sum is -27 -27=2(-11)-5 So to see for usre if this works.. our sum is -27 and that is 5 less than twice -11 which is -22. That is to say -27 is 5 less than -22. Would this work with the positive integers 7, 9, and 11? The sum is 27 and the smallest is 7, but 27 does not equal 14-5, so no.