0.3333
192 divided by 3 is G Than 83 divided by 3. Source: Calculator, Math also Done In Head
"1 divided into g" is written as " g / 1 ".So the question says . . . g/1 = 5/46.And . . . g = 5/46.=============================I think the question actually wanted to say "1 divided BY g", or " 1 / g ".In that case, it says . . . 1/g = 5/46.Multiply each side by g: . . . 1 = 5g/46Multiply each side by 46: . . . 46 = 5gDivide each side by 5: . . . 46/5 = gg = 46/5 = 9.2
You would take the following steps for G = A / (1-R):G = A / (1-R)Multiply by (1-R):G * (1-R) = ADivide by G:(1-R) = A/G1-R = A/GSubtract 1:-R = (A/G) - 1Divide By -1:R = -((A/G) - 1)Check Work:Original Problem:A = 12; R = 5G = 12 / (1-5)G = -3Solving For R:R = -((12/-3)-1)R = 5Therefore, R= -((A/G)-1)
To find the value of G(3), substitute 3 for x in the function G(x) = 4x + 1. This gives G(3) = 4(3) + 1 = 12 + 1 = 13. Therefore, the value of G(3) is 13.
The English word "cabbage" has a letter sum of 100 when assigning values where a=1, b=2, c=3, and so on. The calculation is as follows: c(3) + a(1) + b(2) + b(2) + a(1) + g(7) + e(5) = 3 + 1 + 2 + 2 + 1 + 7 + 5 = 21. However, this does not reach the target sum of 100, indicating that a different approach or word may be needed to achieve that total.
6
1/2(9+p)=p-3
192 divided by 3 is G Than 83 divided by 3. Source: Calculator, Math also Done In Head
"1 divided into g" is written as " g / 1 ".So the question says . . . g/1 = 5/46.And . . . g = 5/46.=============================I think the question actually wanted to say "1 divided BY g", or " 1 / g ".In that case, it says . . . 1/g = 5/46.Multiply each side by g: . . . 1 = 5g/46Multiply each side by 46: . . . 46 = 5gDivide each side by 5: . . . 46/5 = gg = 46/5 = 9.2
library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.STD_LOGIC_ARITH.ALL; use IEEE.STD_LOGIC_UNSIGNED.ALL; ENTITY cldf IS PORT ( a : IN STD_LOGIC_VECTOR(3 DOWNTO 0); b : IN STD_LOGIC_VECTOR(3 DOWNTO 0); ci : IN STD_LOGIC; sum : OUT STD_LOGIC_VECTOR(3 DOWNTO 0); co : OUT STD_LOGI ); END cldf; ARCHITECTURE df OF cldf IS SIGNAL h_sum : STD_LOGIC_VECTOR(3 DOWNTO 0); SIGNAL g : STD_LOGIC_VECTOR(3 DOWNTO 0); SIGNAL p : STD_LOGIC_VECTOR(3 DOWNTO 0); SIGNAL cin : STD_LOGIC_VECTOR(3 DOWNTO 1); BEGIN h_sum <= a XOR b; g <= a AND b; p <= a OR b; PROCESS (g,p,cin) BEGIN cin(1) <= g(0) OR (p(0) AND ci); inst: FOR i IN 1 TO 2 LOOP cin(i+1) <= g(i) OR (p(i) AND cin(i)); END LOOP; co <= g(3) OR (p(3) AND cin(3)); END PROCESS; sum(0) <= h_sum(0) XOR ci; sum(3 DOWNTO 1) <= h_sum(3 DOWNTO 1) XOR cin(3 DOWNTO 1); END df;
5 g is equal to 5000 mg. When divided by 5000 mg, it equals 1.
You would take the following steps for G = A / (1-R):G = A / (1-R)Multiply by (1-R):G * (1-R) = ADivide by G:(1-R) = A/G1-R = A/GSubtract 1:-R = (A/G) - 1Divide By -1:R = -((A/G) - 1)Check Work:Original Problem:A = 12; R = 5G = 12 / (1-5)G = -3Solving For R:R = -((12/-3)-1)R = 5Therefore, R= -((A/G)-1)
3 Parts into which Ancient Gaul was Divided
3
-1
0.1429
To find the value of G(3), substitute 3 for x in the function G(x) = 4x + 1. This gives G(3) = 4(3) + 1 = 12 + 1 = 13. Therefore, the value of G(3) is 13.