Let f ( x ) = 3 x 5 and g ( x ) = 3 x 2 4 x
x+3=14 x+3-3 = 14-3 (subtract 3 from both sides to solve for x) x = 11
-1
d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)] or f'(x) + g'(x) when x = 3, d/dx [f(x) + g(x)] = f'(3) + g'(3) = 1.1 + 7 = 8.1 d/dx [f(x)*g(x)] = f(x)*d/dx[g(x)] + d/dx[f(x)]*g(x) when x = 3, d/dx [f(x)*g(x)] = f(3)*g'(3) + f'(3)*g(3) = 5*7 + 1.1*(-4) = 35 - 4.4 = 31.1
If (x + 3)(x +7) = 0 then either: x + 3 = 0 or x + 7 = 7 Hence x = -3 or -7.
The prime factorization of 312 is 2 X 2 X 2 X 3 X 13In exponential form, 23 x 3 x 132 x 2 x 2 x 3 x 13 = 312
312 = 23 x 3 x 13
2 x 2 x 2 x 3 x 13 = 312 1,2,3,4,6,8,12,13,24,26,39,52,78,104,156,312
104 x 3 = 312
To find the prime factors of 312 start with any factor pair of 312 and keep factoring the composite numbers until all factors are prime. 312 2 x 156 2 x 2 x 78 2 x 2 x 2 x 39 2 x 2 x 2 x 3 x 13 A number of a prime factor of 312 is 2, 3, or 13.
lcm(8 12 13) = 312 Using prime factorisation: 8 = 2³ 12 = 2² x 3 13 = 13 lcm = 2³ x 3 x 13 = 312
The prime factorization of 321 is 3 times 107. Write it any way you wish.
-3(x+5)+3=12
The factors of 312 are 1, 2, 3, 4, 6, 8, 12, 13, 24, 26, 39, 52, 78, 104, 156, 312. For them to be common, they need to be compared to another set of factors. If you meant 3 and 12, the common factors are 1 and 3.
What is the prime factorization of 312. Please tell me don't just gve me an answer that says just joking or something stupid, please with cherry on top!!!!:)
The 10s times table is the easiest one of all! 312 x 10 = 3120 312 x 100 = 31200 312 x 1000 = 312000 312 x 10000 = 3121000
312 x 45 = 14,040