0.3333
the answer to 1 divided by the sum of 3 and g is scrole down plz!
uhh huhh thought i was gonna give the answer huh got fried yupp i knoow you did uhh huhh cant answer that question ahahahahahaha lol thats funny!
6
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)
1
1.07452 + 4.089 = 5.16352 g
6
192 divided by 3 is G Than 83 divided by 3. Source: Calculator, Math also Done In Head
1/2(9+p)=p-3
"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;
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
3 Parts into which Ancient Gaul was Divided
I assume this is the reaction you are talking about:CH4 + H2O (g) --> CO(g) + 3H2(g)From thermodynamics you can approximate the standard heat of reaction Hrxn with Hess' Law. This is the sum of the heats of formations of the products minus the sum of the heats of formations of the reactants in their stoichemtric ratios.Hrxn = SUM [(3)*Hf (H2)+(1)*Hf(CO)] - SUM [(1)*Hf (CH4) + (1)*Hf(H2O (g))]Looking up these values in a Chemical Engineering Handbook or textbook we can substitue and find the heat of reaction.Hrxn = SUM [(3)*(0 kJ/mol) + (1)*(-110.52 kJ/mol)] - SUM [(1)*(-74.85 kJ/mol) +(1)*(-241.83 kJ/mol)]= -110.52 - (-316.68) kJ/mol= + 206.16 kJ/mol, therefore the reaction is endothermicRemember standard heats of formation of elements, as hydrogen in this case, are zero.
-1
0.1429
91 divided by 6 is 15 with remainder 1.