C Language ProblemPlease Explain If int n1 = 5, and int d1 = 2, wh.pdf

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C Language Problem Please Explain If int n1 = 5, and int d1 = 2, what are the results of the following operations? (assume double frac) frac = (double)n1/(double)d1;frac = (double)n1/d1 + 3.5; frac = (double)(n1/d1) + 2.5; Solution Here the heart of functionality is typecasting. What is typecasting? Type casting is a way to convert a variable from one data type to another data type. frac = (double)n1/(double)d1; n1 = 5 and d1 = 2 what this (double)n1/(double)d1 does.. here typecasting is preforming so (double)n1 = 5.000000 (double)d1 = 2.000000 frac = 5.000000/2.000000 frac = 2.500000 frac = (double)n1/d1 + 3.5; Here only n1 is typecasted into double so n1 is 5.000000 and d1 is 2 (double)n1/d1 + 3.5 => (5.000000 / 2) + 3.5 => 2.500000 + 3.5 => 6.000000 frac is 6.000000 frac = (double)(n1/d1) + 3.5; here there is no typecasting between n1 and d1 ... after division is perform then typecasting is going to applied so (double)(n1/d1) + 3.5 (double)(5/2) + 3.5 (double)(2) + 3.5 2.000000 + 3.5 5.500000 Answer is 5.500000.

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$C Language Problem Please Explain If int n1 = 5, and int d1 = 2, what are the results of the following operations? (assume double frac) frac = (double)n1/(double)d1;frac = (double)n1/d1 + 3.5; frac = (double)(n1/d1) + 2.5; Solution Here the heart of functionality is typecasting. What is typecasting? Type casting is a way to convert a variable from one data type to another data type. frac = (double)n1/(double)d1; n1 = 5 and d1 = 2 what this (double)n1/(double)d1 does.. here typecasting is preforming so (double)n1 = 5.000000 (double)d1 = 2.000000 frac = 5.000000/2.000000 frac = 2.500000 frac = (double)n1/d1 + 3.5; Here only n1 is typecasted into double so n1 is 5.000000 and d1 is 2 (double)n1/d1 + 3.5 => (5.000000 / 2) + 3.5 => 2.500000 + 3.5 => 6.000000 frac is 6.000000 frac = (double)(n1/d1) + 3.5; here there is no typecasting between n1 and d1 ... after division is perform then typecasting is going to applied so (double)(n1/d1) + 3.5 (double)(5/2) + 3.5 (double)(2) + 3.5 2.000000 + 3.5 5.500000 Answer is 5.500000$

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most extra cellular environments contain a high concentration of sodium ions (Na+). consider a cell membrane that is permeable to potassium (K+) and is in equilibrium. the diffusion of K+ has left the inside of the ell slightly negatively charged and the outside slightly positively charged. suppose that the cell membrane now opens some sodium channels what happens? A) there is no flow across the membrane. the outside of the cell is positeely charged drivingthe sodium ions father away from the cell. B) there is no flow across the membrane. the cell is in equilibrim and so the electric repulsionof the Na+ exactly counters the effects of diffusion. C) the cell becomes positvely charged. both diffusion and the electric attraction to the cell drive into the cell D) the cell becomes positvely charged; diffusion is reversed and K+ are drived back into the cell Solution The cell becomes positively charged, diffusion is reversed and K+ are drived back into the cell..

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C Language ProblemPlease Explain If int n1 = 5, and int d1 = 2, wh.pdf