In the expression (12a^3 + 16a + 4), the coefficients are the numerical factors that multiply each term. The coefficient of (a^3) is 12, the coefficient of (a) is 16, and the constant term (which can be considered as (a^0)) has a coefficient of 4. Therefore, the coefficients are 12, 16, and 4.
They are 3 and 4
-5 & 3
-4 + 2A = 12 - 15A + A-4 + 2A = 12 - 16A2A = 12 - 16A + 42A = 16 - 16AA = 8 - 8A
2w-6w+9 -4w+9 the coefficient of -4w is -4
To divide ( 48a^3 + 32a^2 + 16a ) by ( 4a ), you can divide each term individually. This gives: ( \frac{48a^3}{4a} = 12a^2 ) ( \frac{32a^2}{4a} = 8a ) ( \frac{16a}{4a} = 4 ) Thus, the result is ( 12a^2 + 8a + 4 ).
(48a3+32a2+16a)/4a = 12a2+8a+4
It is: 19a+27b-9 simplified
1,4,6,4,1
They are 3 and 4
The sum of the coefficients of the chemical equation SF4 + H2O + H2SO + HF is 1 + 2 + 1 + 1 = 5.
-5 & 3
Assuming that "terma coffients" stands for terms and coefficients, they areterms: 3d, d4 coefficients: 3, 4.
2, 4, 4.5 and 17
-4 + 2A = 12 - 15A + A-4 + 2A = 12 - 16A2A = 12 - 16A + 42A = 16 - 16AA = 8 - 8A
You need to combine the different parts that contain the variable "a". Just add the coefficients together algebraically.
2w-6w+9 -4w+9 the coefficient of -4w is -4
16A