Calculations for Veterinary Nurses by Margaret C. Moore, Norman G. Palmer

By Margaret C. Moore, Norman G. Palmer

This useful notebook may also help veterinary nurses with all kinds of calculations. a variety of labored examples are incorporated to improve the reader's self belief in undertaking the techniques concerned. each one form of calculation has its personal separate part within the publication and the authors have used the best attainable technique in explaining each. The publication is dependent in one of these method that the reader can development from an easy rationalization of the mathematics rules concerned, to the appliance of those ideas to crucial veterinary calculations.

Qualified veterinary nurses and scholars alike will locate this booklet a useful reference resource, no matter if acting proper veterinary calculations or learning for pro examinations.

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The only place it can go is to the other side of the equation. Taking the B to the other side involves changing its sign from ‡ B to ÀB. Therefore A ˆ C À B. Again, this can be proven by replacing the letters with the same numbers that were used in the previous examples. Therefore 2 ˆ 6 À 4 Rule 3 Fractional equations can be cross-multiplied. A C Fractional equations such as ˆ can be simpli¢ed using a B D technique known as cross-multiplication. This involves multiplying the bottom of the left hand side by the top of the right hand side and multiplying the bottom of the right hand side by the top of the left hand side.

5 g of dextrose? 5  2 ml ˆ 75 ml. 5 g of dextrose. 5 g of solute in 750 ml of solution. 25% strength solution. This requires the weight to volume ratio of the above solution to be calculated. 25% concentration. % solution ˆ Two sets of data have now been calculated from the data given in the question. 5% solution contain 1250 mg of solute. ' In order to compare the two solutions, either the weights or the volumes need to be equal. e. volume added by the addition of more water, the weight needs to remain constant.

5% solution. 100 ml of a 50% solution contains 50 g of drug. 5% solution. 1 That is only 2X5 50 or 20 of the amount. 1 1 20 of the drug will be contained in 20 of the volume of the 50% solution. 5 g of drug will be 100 20 ˆ 5 ml. e. 5% solution of this drug. 5 % solution, take 25 ml of the 50% solution and make it up to 500 ml with 475 ml of sterile water. Example 3 Once again a 50% solution is available. This time, 500 ml of a 10% solution is required from it. What needs to be done to complete this task?

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