If you cards are only 5,6,7, then the probability of getting even, then even would be zero.
The directions say that the card is not replaced. Therefore, once you pick the 6, there would be no more even cards.
75 units squared. separate the shapes into a rectangle and triangle, find the area of each, and add together
If x=-5 is a zero, then the first factor of the polynomial would be (x + 5 )
To find the other two factors we can divide the polynomial by the expression (x+5).
Using synthetic division, we have:
-5 I 4 15 -24 5 (Coefficients of the dividend)
I -20 25 -5 (Multiplying each coefficient by the results of the substraction and adding)
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4 -5 1 0 (Coefficients of the quotient)
The result of the division is 4x^2 - 5x + 1. Factoring it, we have:
4x^2 - 4x -x + 1 (Separating -5x into -x and -4x)
4x (x - 1) - (x -1) (Factoring each pair of terms)
(x-1)(4x-1) (Factoring using the common factor)
So the answer would be:
(x + 5 )(x-1)(4x-1)
Answer: cos(x)
Step-by-step explanation:
We have
sin ( x + y ) = sin(x)*cos(y) + cos(x)*sin(y) (1) and
cos ( x + y ) = cos(x)*cos(y) - sin(x)*sin(y) (2)
From eq. (1)
if x = y
sin ( x + x ) = sin(x)*cos(x) + cos(x)*sin(x) ⇒ sin(2x) = 2sin(x)cos(x)
From eq. 2
If x = y
cos ( x + x ) = cos(x)*cos(x) - sin(x)*sin(x) ⇒ cos²(x) - sin²(x)
cos (2x) = cos²(x) - sin²(x)
Hence:The expression:
cos(2x) cos(x) + sin(2x) sin(x) (3)
Subtition of sin(2x) and cos(2x) in eq. 3
[cos²(x)-sin²(x)]*cos(x) + [(2sen(x)cos(x)]*sin(x)
and operating
cos³(x) - sin²(x)cos(x) + 2sin²(x)cos(x) = cos³(x) + sin²(x)cos(x)
cos (x) [ cos²(x) + sin²(x) ] = cos(x)
since cos²(x) + sin²(x) = 1
