Anything times zero is zero.
6 x 0 x 5 = 0 (6 x 0) = 0 (0 x 5) = 0 So your answer is zero.
6 x 0 x 5 = 0 Because any number multiplied by 0 is equal to zero, 6 x 0 = 0 and then 0 x 5 = 0.
x^2 + 12x + 36 = 0(x + 6)(x + 6) = 0(x + 6)^2 = 0√(x + 6)^2 = ± √0x + 6 = 0x + 6 - 6 = 0 - 6x = 6
x = 12
cos x - 2 sin x cos x = 0 -> cos x (1 - 2 sin x) = 0 => cos x = 0 or 1 - 2 sin x = 0 cos x = 0: x = π/2 + kπ 1 - 2 sin x = 0: sin x = 1/2 -> x = π/6 + 2kπ or 5/6π + 2kπ Thus x = π/2 + kπ; x = π/6 + 2kπ; x = 5/6π + 2kπ solve the original equation.
x=-6+6=0 then x=-2+2=0 and 0+0=0
This is relatively easy to solve, only requires application of the laws of indices. (x+5)1/5 - (x+5)6/5 = 0 => (x+5)1/5 - (x+5)(x+5)1/5 = 0 => (x+5)1/5 (1 - (x+5)) = 0 => -(x+5)1/5 (x+4)=0 since the expression equals 0 one term or the other has to be 0 therefore x= -4 or x= -5
5 ^x = 0 5 = root square 0 5=0
Find two numbers that sum to -6 and which multiply to 5. The answer to that is -1 and -5 So: x2 - 6x + 5 = 0 implies (x - 1)*(x - 5) = 0 which is true if and only if x - 1 = 0 or x - 5 = 0 which is true if and only if x = 1 or x = 5
it is 0 because 6 x 0 =0 then 0 x 5 =0
6 x 0 x 5 = 0 (6 x 0) = 0 (0 x 5) = 0 So your answer is zero.
To solve the equation (5 \cos(x)^2 + 8 \cdot 5 \cos(x) = 0), first factor out (5 \cos(x)): [5 \cos(x)(\cos(x) + 8) = 0.] This gives two cases to solve: (5 \cos(x) = 0) (which leads to (\cos(x) = 0)) and (\cos(x) + 8 = 0) (which is not possible since the range of (\cos(x)) is ([-1, 1])). Therefore, solve (\cos(x) = 0) to find (x = \frac{\pi}{2} + n\pi), where (n) is any integer.
6 x 0 x 5 = 0 Because any number multiplied by 0 is equal to zero, 6 x 0 = 0 and then 0 x 5 = 0.
x^2 + 12x + 36 = 0(x + 6)(x + 6) = 0(x + 6)^2 = 0√(x + 6)^2 = ± √0x + 6 = 0x + 6 - 6 = 0 - 6x = 6
To solve this equation we have to isolate and solve for x. So: 2x-6= x+5 +6 +6 2x = x+11 -x -x x = 11 So x=11.
factorise (x-5)(x+4)=0 Hence x= 5 or -4
X squared - 3x - 10 = 0 (x-5) (x+2) = 0 x - 5 = 0 x + 2 = 0 X= 5 x = -2