What Are the Fees of Java in ABC for Technology Training?

I think its 24000 Including Java & Software Testing

1. ABC is a three-digit number where A>0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?

The ABC is a three digit number where A>0 . The ABC is equal so B>0

2. When will FMS DU be at par with IIM ABC?

Thanks for A2AI think FMS DU is at par with IIM ABC.There are various standards on which we can compare both.1.) FeesHere FMS DU is way above than IIMs. FMS charges less than 50k where IIMs chareges approx 20lakh. Peer Group. Both group have students from various streams, personalities, way of thinking and non other than CAT score (99.5 percentile for general).PlacementsBoth the groups have the place range of 15-25 lakh with average around 20lakhs.CampusWhile the IIM ABC has a very large campus FMS is a 2 classroom building, but FMS DU can avail the facilities of DU like library, sports Complex. AreaAll these Colleges are in the metropolitan cities. So everyone is equal here also.Learning ExperienceIt depends on which mindset a student enters in the college. So it is on the student how they grab the opportunity available for him. AlumniIIM ABC and FMS has their alumni in every sector but due to large number of student in IIM's they are slightly better in number here.Raavan

3. The number of all 3-digit numbers abc for which abc abbc acabc = 29

$$(a1)(b1)(c1)=30$$Total number of ways to satisfy above equation:In the first case we have $6$ values for $a,b,c$ as $(1,2,4)$ can take up any digit hence they can be arranged in $3!$ ways.In the second case we have $4$ values for $a,b,c$ as $(0,5,4)$. $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number ( contrary to given condition ). Hence we have $ 2. 2. 1$ number of ways . Similarly in the third case we have $4$ values $a,b,c$ as $(0,2,9)$. $0$ can not be as first digit as. Then $abc$ would be a $2$ digit number ( contrary to given condition ). Hence we have $ 2.2.1$ number of ways .Therefore, total number of ways:$$3!2times 2times12times 2times1=644=14 space textways$$.

4. If ABC is an acute triangle, how do I prove that the area of the orthic triangle of ABC is

Since angle BPA = 90^circ, we have BP = AB * cos(angle B). Similarly, BR = BC * cos(angle B). Therefore, if mathcalA denotes the area of a triangle, then we have:beginarraycclmathcalA_BPR & = & BR * BP * dfracsin(angle B)2 & = & AB * BC * dfracsin(angle B)2 * cos^2(angle B) & = & mathcalA_ABC * cos^2(angle B) endarraySimilarly, mathcalA_ARQ = mathcalA_ABC * cos^2(angle A) and mathcalA_CPQ = mathcalA_ABC * cos^2(angle C)Now since triangle ABC is acute, the area of the triangle we want is the area of triangle ABC minus the areas of triangle BPR, triangle ARQ and triangle CPQ. Therefore, the area of the orthic triangle is mathcalA_PQR = mathcalA_ABC left(1 - cos^2(angle A) - cos^2(angle B) - cos^2(angle C)right)We can show using trigonometric identities that since angle A angle B angle C = 180^circ, then 1 - cos^2(angle A) - cos^2(angle B) - cos^2(angle C) = 2cos(angle A)cos(angle B)cos(angle C). Also, there is a formula which says mathcalA_ABC = 2R^2 sin(angle A)sin(angle B)sin(angle C), where R is the circumradius of triangle ABC. I'll leave the proof of these two theorems as an exercise.Therefore, substituting these two equations into our equation for the area of the orthic triangle gives us the following:beginarrayccl mathcalA_PQR & = & 2R^2 sin(angle A)sin(angle B)sin(angle C) * 2cos(angle A)cos(angle B)cos(angle C) & = & dfracR^2 sin(2A)sin(2B)sin(2C)2 endarrayThat's the simplest proof I could come up with. frac R^2sin2Acdotsin2Bcdotsin2C2?If ABC is an acute traingle, how do I prove that the area of the orthic triangle of ABC is frac R^2sin2Acdotsin2Bcdotsin2C2?

5. In triangle ABC, according to the normal symbols,

textLet,,dfraca b5 = dfracb c7 = dfracc a6 = mimplies a b = 5 mqquad b c = 7 mqquad c a = 6 mtherefore ,,(c a) - (a b) = 6 m - 5 mimplies c - b = mimplies c = b mtag 1(b c) - (c a) = 7 m - 6 mimplies b - a = mimplies b = a mtag 2textby sine lawdfracsin,Aa = dfracsin,Bb = dfracsin,CctextLet,,dfracsin,Aa = dfracsin,Bb = dfracsin,Cc = ktherefore,,sin,A = a ksin,B = b k = (a m) k = a k mkquad textby (2)sin,C = c k = (b m) k = b k m k = (a m) k m k = a 2 m kquad textby (1) & (2)sin, A = akqquad sin,B = a k m kqquad sin,C = ak 2 m ktextIt is an arithmetic sequence with first term ak and common difference mk.fracab5=fracbc7=fracca6. How do you prove that sinA,sinB,sinC is an arithmetic sequence?