Expressed values are considered significant to the last digit shown (seeSignificant Figures and Tolerancesin theGeneral Notices).Significant figures are digits with practical meaning.The accuracy of the determination is implied by the number of figures used in its expression.In some calculations zeros may not be significant.For example,for a measured weight of 0.0298g,the zeros are not significant;they are used merely to locate the decimal point.In the example,2980g,the zero may also be used to indicate the decimal point,in which case the zero is not significant.Alternately,however,the zero may indicate that the weight is closer to 2981g or 2979g,in which case the zero is significant.In such a case,knowledge of the method of measurement would be required in order to indicate whether the zero is or is not significant.In the case of a volume measurement of 298mL,all of the digits are significant.In a given result,the last significant figure written is approximate but all preceding figures are accurate.For example,a volume of 29.8mLimplies that 8is approximate.The true volume falls between 29.75and 29.85.Thus,29.8mLis accurate to the nearest 0.1mL,which means that the measurement has been made within ±0.05mL.Likewise,a value of 298mLis accurate to the nearest 1mLand implies a measurement falling between 297.5and 298.5,which means that the measurement has been made within ±0.5mLand is subject to a maximum error calculated as follows:

Azero in a quantity such as 298.0mLis a significant figure and implies that the measurement has been made within the limits of 297.95and 298.05with a possible error calculated as follows:

1. 29.8mL=29.8±0.05mL(accurate to the nearest 0.1mL)
2. 29.80mL=29.80±0.005mL(accurate to the nearest 0.01mL)
3. 29.800mL=29.800±0.0005mL(accurate to the nearest 0.001mL)

Measurement	Number of Significant Figures
2.98	3
2.980	4
0.0298	3
0.0029	2

Sums and Differences— When adding or subtracting,the number of decimal places in the result shall be the same as the number of decimal places in the component with the fewest decimal places.

Products and Quotients— When multiplying or dividing,the result shall have no more significant figures than the measurement with the smallest number of significant figures entering into the calculation.

The logarithm of a number is the exponent or the power to which a given base must be raised in order to equal that number.

pH=–log [H+],and

pKa =–log Ka

pH=pKa +log [salt]/[acid]

The pharmacist must be able to calculate the amount or concentration of drug substances in each unit or dosage portion of a compounded preparation at the time it is dispensed.Pharmacists must perform calculations and measurements to obtain,theoretically,100%of the amount of each ingredient in compounded formulations.Calculations must account for the active ingredient,or active moiety,and water content of drug substances,which includes that in the chemical formulas of hydrates.Official drug substances and added substances must meet the requirements underLoss on Drying á731ñ,which must be included in the calculations of amounts and concentrations of ingredients.The pharmacist should consider the effect of ambient humidity on the gain or loss of water from drugs and added substances in containers subjected to intermittent opening over prolonged storage.Each container should be opened for the shortest duration necessary and then closed tightly immediately after use.

The nature of the drug substance that is to be weighed and used in compounding a prescription must be known exactly.If the substance is a hydrate,its anhydrous equivalent weight may need to be calculated.On the other hand,if there is adsorbed moisture present that is either specified on a certificate of analysis or that is determined in the pharmacy immediately before the drug substance is used by the procedure underLoss on Drying á731ñ,this information must be used when calculating the amount of drug substance that is to be weighed in order to determine the exact amount of anhydrous drug substance required.

There are cases in which the required amount of a dose is specified in terms of a cation [e.g.,Li+,netilmicin (n+)],an anion [e.g.,F–],or a molecule (e.g.,theophylline in aminophylline).In these instances,the drug substance weighed is a salt or complex,a portion of which represents the pharmacologically active moiety.Thus,the exact amount of such substances weighed must be calculated on the basis of the required quantity of the pharmacological moiety.

The following formula may be used to calculate the exact theoretical weight of an ingredient in a compounded preparation:

in whichWis the actual weighed amount;ais the prescribed or pharmacist-determined weight of the active or functional moiety of drug or added substance;bis the chemical formula weight of the ingredient,including waters of hydration for hydrous ingredients;dis the fraction of dry weight when the percent by weight of adsorbed moisture content is known from the loss on drying procedure (seeLoss on Drying á731ñ);and eis the formula weight of the active or functional moiety of a drug or added substance that is provided in the formula weight of the weighed ingredient.

Equation Factor	Numerical Value
W	weight,in g,of Morphine Sulfate USP
a	1.5g of morphine sulfate pentahydrate in the prescription
b	759g/mole
d	1.0
e	759g/mole

Equation Factor	Numerical Value
W	weight,in mg,of Aminophylline USP (dihydrate)
a	250mg of theophylline
b	456g/mole
d	0.996
e	360g/mole

Equation Factor	Numerical Value
W	weight,in g,of Lithium Citrate USP (tetrahydrate)
a	200mEq of Li+or 1.39g of Li+
b	282g/mole
d	0.975
e	3×6.94g/mole or 20.8g/mole

Equation Factor	Numerical Value
W	weight,in g,of Netilmicin Sulfate USP
a	2.5g
b	1442g/mole
d	0.88
e	951g/mole

EXAMPLE—

(mg of drug per kg of body weight )×(kg of body weight)=dose for an individual for a 24-hour period

(amount of drug per m2of body surface area)×(body surface area in m2)=dose for an individual for a 24-hour period.

BSA(m2)=square root of {[Height (in)×Weight (lb)]/3131}

BSA(m2)=square root of {[Height (cm)×Weight (kg)]/3600}.

EXAMPLE—

Percentage concentrations of solutions are usually expressed in one of three common forms:

See alsoPercentage Measurementsunder Concentrationsin the General Notices.The above three equations may be used to calculate any one of the three values (i.e.,weights,volumes,or percentages)in a given equation if the other two values are known.

Note that weights are always additive,i.e.,50g plus 25g =75g.Volumes of two different solvents or volumes of solvent plus a solid solute are not strictly additive.Thus 50mLof water +50mLof pure alcohol do not produce a volume of 100mL.Nevertheless,it is assumed that in some pharmaceutical calculations,volumes are additive,as discussed below underReconstitution of Drugs Using Volumes Other than Those on the Label.

1. Calculate the percentage concentrations (w/w)of the constituents of the solution prepared by dissolving 2.50g of phenol in 10.00g of glycerin.Using the weight percent equation above,the calculation is as follows.
   Total weight of the solution =10.00g +2.50g =12.50g.
   Weight percent of phenol =(2.50g ×100%)/12.50g =20.0%of phenol.
   Weight percent of glycerin =(10g ×100%)/12.50g =80.0%of glycerin.
2. Aprescription order reads as follows:
   Eucalyptus Oil 3%(v/v)in Mineral Oil.
   Dispense 30.0mL.
   What quantities should be used for this prescription?Using the volume percent equation above,the calculation is as follows.
   Amount of Eucalyptus Oil:
   3%=(Volume of oil in mL/30.0mL)×100%
   Solving the equation,the volume of oil =0.90mL
   Amount of Mineral Oil:
   To 0.90mLof Eucalyptus Oil add sufficient Mineral Oil to prepare 30.0mL.
3. Aprescription order reads as follows:
   Zinc oxide	7.5g
   Calamine	7.5g
   Starch	15g
   White petrolatum	30g

   Calculate the percentage concentration for each of the four components.Using the weight percent equation above,the calculation is as follows.
   Total weight =7.5g +7.5g +15g +30g =60.0g.
   Weight percent of zinc oxide =(7.5g zinc oxide/60g ointment)×100%=12.5%.
   Weight percent of calamine =(7.5g calamine/60g ointment)×100%=12.5%.
   Weight percent of starch =(15g starch/60g ointment)×100%=25%.
   Weight percent of white petrolatum =(30g white petrolatum/60g ointment)×100%=50%.

The definition of Specific Gravity is usually based on the ratio of weight of a substance in air at 25to that of the weight of an equal volume of water at the same temperature.The weight of 1mLof water at 25is approximately 1g.The following equation may be used for calculations.

1. Aliquid weighs 125g and has a volume of 110mL.What is the specific gravity?
   The weight of an equal volume of water is 110g.
   Using the above equation,specific gravity =125g/110g =1.14.
2. Hydrochloric Acid NFis approximately 37%(w/w)solution of hydrochloric acid (HCl)in water.How many grams of HCl are contained in 75.0mLof HCl NF?(Specific gravity of Hydrochloric Acid NFis 1.18.)
   Calculate the weight of HCl NFusing the above equation.
   The weight of an equal volume of water is 75g.
   Specific Gravity 1.18=weight of the HCl NFg /75.0g.
   Solving the equation,the weight of HCl NFis 88.5g.
   Now calculate the weight of HCl using the weight percent equation.
   37.0%(w/w)=(weight of solute g/88.5g )×100.
   Solving the equation,the weight of the HCl is 32.7g.

Aconcentrated solution can be diluted.Powders and other solid mixtures can be triturated or diluted to yield less concentrated forms.Because the amount of solute in the diluted solution or mixture is the same as the amount in the concentrated solution or mixture,the following relationship applies to dilution problems.

Almost any quantity and concentration terms may be used.However,the units of the terms must be the same on both sides of the equation.

1. Calculate the amount (Q2),in g,of diluent that must be added to 60g of a 10%(w/w)ointment to make a 5%(w/w)ointment.
   Let (Q1)=60g,(C1)=10%,and (C2)=5%.
   Using the above equation,
   60g ×10%=(Q2)×5%(w/w)
   Solving the above equation,the amount of product needed,Q2,is 120g.The initial amount of product added was 60g,and therefore an additional 60g of diluent must be added to the initial amount to give a total of 120g.
2. How much diluent should be added to 10g of a trituration (1in 100)to make a mixture that contains 1mg of drug in each 10g of final mixture?
   Determine the final concentration by first converting mg to g.One mg of drug in 10g of mixture is the same as 0.001g in 10g.
   Let (Q1)=10g,(C1)=(1in 100),and (C2)=(0.001in 10).
   Using the equation for dilution,10g ×(1/100)=(Q2)g ×(0.001/10).
   Solving the above equation,(Q2)=1000g.
   Because 10g of the final mixture contains all of the drug and some diluent,(1000g –10g)or 990g of diluent is required to prepare the mixture at a concentration of 0.001g of drug in 10g of final mixture.
3. Calculate the percentage strength of a solution obtained by diluting 400mLof a 5.0%solution to 800mL.
   Let (Q1)=400mL,(C1)=5%,and (Q2)=800mL.
   Using the equation for dilution,400mL×5%=800mL×(C2)%.
   Solving the above equation,(C2)=2.5%(w/v).

See Units of Potencyin the General Notices.

Because some substances may not be able to be defined by chemical and physical means,it may be necessary to express quantities of activity in biological units of potency.

1. One mg of Pancreatin contains not less than 25USP Units of amylase activity,2.0USP Units of lipase activity,and 25USP Units of protease activity.If the patient takes 0.1g (100mg)per day,what is the daily amylase activity ingested?
   1mg of Pancreatin corresponds to 25USP Units of amylase activity.
   100mg of Pancreatin corresponds to 100×(25USP Units of amylase activity)=2500Units.
2. Adose of penicillin Gbenzathine for streptococcal infection is 1.2million units intramuscularly.If a specific product contains 1180units per mg,how many milligrams would be in the dose?
   1180units of penicillin Gbenzathine are contained in 1mg.
   1unit is contained in 1/1180mg.
   1,200,000units are contained in (1,200,000×1)/1180units =1017mg.

Frequently the base form of a drug is administered in an altered form such as an ester or salt for stability or other reasons such as taste or solubility.This altered form of the drug usually has a different molecular weight (MW),and at times it may be useful to determine the amount of the base form of the drug in the altered form.

1. Four hundred milligrams of erythromycin ethylsuccinate (molecular weight,862.1)is administered.Determine the amount of erythromycin (molecular weight,733.9)in this dose.
   862.1g of erythromycin ethylsuccinate corresponds to 733.9g of erythromycin.
   1g of erythromycin ethylsuccinate corresponds to (733.9/862.1)g of erythromycin.
   0.400g erythromycin ethylsuccinate corresponds to (733.9/862.1)×0.400g or 0.3405g of erythromycin.
2. The molecular weight of testosterone cypionate is 412.6and that of testosterone is 288.4.What is the dose of testosterone cypionate that would be equivalent to 60.0mg of testosterone?
   288.4g of testosterone corresponds to 412.6g of testosterone cypionate.
   1g of testosterone corresponds to 412.6/288.4g of testosterone cypionate.
   60.0mg or 0.0600g of testosterone corresponds to (412.6/288.4)×0.0600=0.0858g or 85.8mg of testosterone cypionate.

Occasionally it may be necessary to reconstitute a powder in order to provide a suitable drug concentration in the final product.This may be accomplished by estimating the volume of the powder and liquid medium required.

1. If the volume of 250mg of ceftriaxone sodium is 0.1mL,how much diluent should be added to 500mg of ceftriaxone sodium powder to make a suspension having a concentration of 250mg per mL?
   
2. Volume of 500mg of ceftriaxone sodium =
   
3. Volume of the diluent required =(2mLof suspension)-(0.2mLof Ceftriaxone Sodium)=1.8mL.
4. What is the volume of dry powder cefonicid,if 2.50mLof diluent is added to 1g of powder to make a solution having a concentration of 325mg per mL?
   Volume of solution containing 1g of the powder =

   
   Volume of dry powder cefonicid =3.08mLof solution -2.50mLof diluent =0.58mL.

1. Place the desired percentage or concentration in the center.
2. Place the percentage of the substance with the lower strength on the lower left-hand side.
3. Place the percentage of the substance with the higher strength on the upper left-hand side.
4. Subtract the desired percentage from the lower percentage,and place the obtained difference on the upper right-hand side.
5. Subtract the higher percentage from the desired percentage,and place the obtained difference on the lower right-hand side.

EXAMPLES—

EXAMPLES—

See Concentrationsin the General Notices.

EXAMPLES—

The quantities of electrolytes administered to patients are usually expressed in terms of mEq.This term must not be confused with a similar term used in quantitative chemical analysis as discussed above.Weight units such as mg or g are not often used for electrolytes because the electrical properties of ions are best expressed as mEq.An equivalent is the weight of a substance (equivalent weight)that supplies one unit of charge.An equivalent weight is the weight,in g,of an atom or radical divided by the valence of the atom or radical.Amilliequivalent is one-thousandth of an equivalent (Eq).Because the ionization of phosphate depends on several factors,the concentration is usually expressed in millimoles,moles,or milliosmoles which are described below.[NOTE—Equivalent weight (Eq.wt)=wt.of an atom or radical (ion)in g/valence (or charge)of the atom or radical.Milliequivalent weight (mEq.wt)=Eq.wt.(g)/1000.]

1. Potassium (K+)has a gram-atomic weight of 39.10.The valence of K+is 1+.Calculate its milliequivalent weight (mEq wt).
   Eq wt =39.10g/1=39.10g
   mEq wt =39.10g/1000=0.03910g =39.10mg
2. Calcium (Ca2+)has a gram-atomic weight of 40.08.Calculate its milliequivalent weight (mEq wt).
   Eq wt =40.08g/2=20.04g
   mEq wt.=20.04g/1000=0.02004g =20.04mg
   NOTE—The equivalent weight of a compound may be determined by dividing the molecular weight in g by the product of the valence of either relevant ion and the number of times this ion occurs in one molecule of the compound.
3. How many milliequivalents of potassium ion (K+)are there in a 250-mg Penicillin V Potassium Tablet?[NOTE—Molecular weight of penicillin Vpotassium is 388.48g per mol;there is one potassium atom in the molecule;and the valence of K+is 1.]
   Eq wt =388.48g/[1(valence)×1(number of charges)]=388.48g.
   mEq wt =388.48g/1000=0.38848g =388.48mg.
   (250mg per Tablet)/(388.48mg per mEq)=0.644mEq of K+per Tablet.
4. How many equivalents of magnesium ion and sulfate ion are contained in 2mLof a 50%Magnesium Sulfate Injection?(Molecular weight of MgSO4·7H2Ois 246.48g per mol.)
   Amount of magnesium sulfate in 2mLof 50%Magnesium Sulfate Injection
   
   Eq wt of MgSO4.7H2O=MW(g)/(valence of specified ion ×number of specified ions in one mole of salt).
   For the magnesium ion:
   The number of equivalents is calculated as follows:
   246.48/[2(valence)×1(number of ions in the compound)]=123.24g/Eq of magnesium ion.
   The number of equivalents in 1g is 1g/123.24g/Eq =0.008114Eq.
   The number of mEq may be calculated as follows.
   The mEq wt =Eq wt (g)/1000=(123.24g/Eq)/1000=0.12324g.
   The number of milliequivalents of magnesium ion in 1g is 1g/0.12324g/mEq =8.114mEq.
   For the sulfate ion:
   The number of equivalents is calculated as follows:
   246.48/[2(valence)×1(number of ions in the compound)]=123.24g/Eq of sulfate ion.
   The number of equivalents in 1g is 1g/123.24g/Eq =0.008114Eq.
   The number of mEq may be calculated as follows.
   The mEq wt =Eq wt (g)/1000=(123.24g/Eq)/1000=0.12324g.
   The number of milliequivalents of sulfate ion in 1g is 1g/0.12324g/mEq =8.114mEq.
5. Avial of Sodium Chloride Injection contains 3mEq of sodium chloride per mL.What is the percentage strength of this solution?(Molecular weight of sodium chloride is 58.44g per mol.)
   1mEq =1Eq/1000=58.44g/1000=0.05844g =58.44mg.
   Amount of sodium chloride in 3mEq per mL=58.44mg per mEq ×3mEq per mL=175.32mg per mL.
   

EXAMPLES—

The following discussion and calculations have therapeutic implications in preparations of dosage forms intended for ophthalmic,subcutaneous,intravenous,intrathecal,and neonatal use.

Cells of the body,such as erythrocytes,will neither swell nor shrink when placed in a solution that is isotonic with the body fluids.However,the measurement of tonicity,a physiological property,is somewhat difficult.It is found that a 0.9%(w/v)solution of sodium chloride,which has a freezing point of –0.52,is isotonic with body fluids and is said to be isoosmotic with body fluids.In contrast to isotonicity,the freezing point depression is a physical property.Thus many solutions which are isoosmotic with body fluids are not necessarily isotonic with body fluids,e.g.,a solution of urea.Nevertheless many pharmaceutical products are prepared using freezing point data or related sodium chloride data to prepare solutions that are isoosmotic with the body fluids.Aclosely related topic is osmolarity (see Osmolality and Osmolarity á785ñ).

Freezing point data or sodium chloride equivalents of pharmaceuticals and excipients (see Table 1below)may be used to prepare isoosmotic solutions,as shown in the examples below.

Table 1.Sodium Chloride Equivalents (E)and Freezing Point (FP)Depressions for a 1%Solution of the Drug or Excipent

Some calculations concerning flow rates in intravenous sets are provided below.[NOTE—Examples below are notto be used for treatment purposes.]

1. Sodium Heparin 8,000units in 250mL Sodium Chloride Injection 0.9%solution are to be infused over 4hours.The administration set delivers 20drops per mL.
   What is the flow rate in mLper hour?
   In 4hours,250mLare to be delivered.
   In 1hour,250mL/4=62.5mLare delivered.
   What is the flow rate in drops per minute?
   In 60minutes,62.5mLare delivered.
   In 1minute,62.5mL/60=1.04mLare delivered.
   1mL=20drops.
   1.04mL=1.04×20drops =20.8drops.
   Thus in 1minute,20.8or 21drops are administered.
2. A14.5kg patient is to receive 50mg of Sodium Nitroprusside in 250mLof dextrose 5%in water (D5W)at the rate of 1.3µg per kg per minute.The set delivers 50drops per mL.
   Calculate the flow rate in mLper hour.
   The dose for 1kg is 1.3µg per minute.
   The 14.5kg patient should receive 14.5×1.3µg =18.85µg per minute.
   50mg or 50000µg of drug are contained in 250mLof D5W.
   18.85µg are contained in 250mL×18.85/50000=0.09425mL D5W,which is administered every minute.
   In 1minute,0.09425mLare administered.
   In 1hour or 60minutes,60×0.09425mL=5.655or 5.7mLare administered.
   Calculate the flow rate in drops per minute.
   1mLcorresponds to 50drops per minute.
   0.09425mLcorresponds to 0.09425×50=4.712or 4.7drops per minute.

The relationship between Celsius degrees (C)and Fahrenheit degrees (F)is expressed by the following equation:

in which Cand Fare the numbers of Celsius degrees and Fahrenheit degrees,respectively.

1. Convert 77Fto Celsius degrees.
   
   9(C)=5(F)–160
   C=[5(F)–160]/9=[(5×77)–160]/9=25C
2. Convert 30Cto Fahrenheit degrees.
   
   9(C)=5(F)–160
   F=[9(C)+160]/5=[(9×30)+160]/5=86F

K=C+273.1,

See Pharmaceutical Stability á1150ñfor the definition of mean kinetic temperature (MKT).MKTis usually higher than the arithmetic mean temperature and is derived from the Arrhenius equation.MKTaddresses temperature fluctuations during the storage period of the product.The mean kinetic temperature,TK,is calculated by the following equation:

in which DHis the heat of activation,which equals 83.144kJper mol (unless more accurate information is available from experimental studies);Ris the universal gas constant,which equals 8.3144×10–3kJper degree per mol;T1is the average temperature,in degrees Kelvin,during the first time period,e.g.,the first week;T2is the average temperature,in degrees Kelvin,during the second time period,e.g.,second week;and Tnis the average temperature,in degrees Kelvin during the nth time period,e.g.,nth week,nbeing the total number of temperatures recorded.The mean kinetic temperature is calculated from average storage temperatures recorded over a one-year period,with a minimum of twelve equally spaced average storage temperature observations being recorded (see Pharmaceutical Stability á1150ñ.This calculation can be performed manually with a pocket calculator or electronically with computer software.

1. The means of the highest and lowest temperatures for 52weeks are 25Ceach.Calculate the MKT.
   n=52
   DH/R=10,000K
   T1,T2,...,Tn=25C=273.1+25=298.1K
   R=0.0083144kJ K–1mol–1
   DH=83.144kJper mol
   
   
   
   
   The calculated MKTis 25.0C.Therefore the controlled room temperature requirement is met by this pharmacy.[NOTE—If the averages of the highest and lowest weekly temperatures differed from each other and were in the allowed range of 15Cto 30C(seeControlled Room TemperatureunderPreservation,Packaging,Storage,and Labelingin the General Notices),then each average would be substituted individually into the equation.The remaining two examples illustrate such calculations,except that the monthly averages are used]
2. Apharmacy recorded a yearly MKTon a monthly basis,starting in January and ending in December.Each month,the pharmacy recorded the monthly highest temperature and the monthly lowest temperature,and the average of the two was calculated and recorded for the MKTcalculation at the end of the year.
   Table 2.Data for Calculation of MKT

   n	Month	Lowest Temperature (in C)	Highest Temperature (in C)	Average Temperature (in C)	Average Temperature (in K)	DH/RT	e–DH/RT
   1	Jan.	15	27	21	294.1	34.002	1.710×10–15
   2	Feb.	20	25	22.5	295.6	33.830	2.033×10–15
   3	Mar.	17	25	21	294.1	34.002	1.710×10–15
   4	Apr.	20	25	22.5	295.6	33.830	2.033×10–15
   5	May	22	27	24.5	297.6	33.602	2.551×10–15
   6	June	15	25	20	293.1	34.118	1.523×10–15
   7	July	20	26	23	296.1	33.772	2.152×10–15
   8	Aug.	22	26	24	297.1	33.659	2.411×10–15
   9	Sept.	23	27	25	298.1	33.546	2.699×10–15
   10	Oct.	20	28	24	297.1	33.659	2.411×10–15
   11	Nov.	20	24	22	295.1	33.887	1.919×10–15
   12	Dec.	22	28	25	298.1	33.546	2.699×10–15

   From these data the MKTmay be estimated or it may be calculated.If more than half of the observed temperatures are lower than 25Cand a mean lower than 23Cis obtained,the MKTmay be estimated without performing the actual calculation.
   To estimate the MKT,the recorded temperatures are evaluated and the average is calculated.In this case,the calculated arithmetic mean is 22.9C.Therefore,the above requirements are met and it can be concluded that the mean kinetic temperature is lower than 25C.Therefore,the controlled room temperature requirement is met.
   The second approach is to perform the actual calculation.
   n=12
   
   
   
   
   The calculated MKTis 23.0C,so the controlled room temperature requirement is met.[NOTE—These data and calculations are used only as an example.]
3. An article was stored for one year in a pharmacy where the observed monthly average of the highest and lowest temperatures was 25C(298.1K),except for one month with an average of 28C(301.1K).Calculate the MKTof the pharmacy.
   
   

   n=12
   
   
   
   
   
   
   
   The controlled room temperature requirement is not met because the calculated MKTexceeds 25C.(See Notein Example 2above.)