Using Pascal's Triangle
C( n , k) + C(n, k + 1) = C(n+1, k + 1)
So
C(26, 13) + C(26, 14) = C( 27, 14)
But ....C(27, 14) = C(27, 13) ....so we also have
C(26, 12) + C(26, 13) = C(27, 13)
n = 12 or n = 14
And their sum = 26
< 58.5 < 63.5 < 68.5
44 77 80
If 4 + i a root then so is the conjugate 4 - i
Unfortunately.....We have to multiply these....!!!!
Multiplying these we get the polynomial
(x - (4 + i) ) ( x - (4 - i) ) = x^2 - x(4 + i) - x(4 - i) + (4 + i)(4-i) =
x^2 - 8x + 16 - i^2 =
x^2 - 8x + 17
To find the remaining polynomial....we can perform some polynomial division
x + 3
x^2 - 8x + 17 [ x^3 - 5x^2 - 7x + 51 ]
x^3 - 8x^2 + 17x
___________________
3x^2 - 24x + 51
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Since x + 3 is the remaining polynomial, x = -3 is the remaining root
Answered here, too :
https://web2.0calc.com/questions/help-urgent_19
The number of data values is [ 10 * 11] / 2 = 110/2 = 55
The mean is
[1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + + 8^2 + 9^2 + 10^2] / 55 = 7
The mode is the data value that appears most often = 10
The median is also 7
RPQ is supplemental to RPS
And the sines of supplemental angles are equal
Thus...... sin RPQ = sin RPS = 7/25
And since RPS is obtuse, the cosine of this angle = - sqrt[ 25^2 - 7^2 ] / 25 =
- sqrt [ 625 - 49 ] / 25 =
- sqrt [ 576] / 25 =
-24 / 25
[ ( 4 - (-5)]^2 = [ 4 + 5]^2 = 9^2 = 81 (1)
[ ( -9 - 2 ]^2 = [ -11]^2 = 121 (2)
Add (1) and (2), NSS
Take the square root of this sum and that will be your answer
Excellent, NSS !!!!
Adjacent side to R = 12
Hypotenuse = 13
Do you see the answer ???
Opposite side to R = 5
