Explore topic-wise InterviewSolutions in Current Affairs.

ABC/Step-by-step EXPLANATION:jikwn N o kjwowoowwhjqoqqkwowieb JWO o

Step-by-step EXPLANATION:""8,! Bhichr MAZ PJ

for FENCING we have to FIND perimeterperimeter=2(l+b)=2(10+15)=2*25=50mCost of fencing=ruppee 30*50=1500ruppee

l = 30 Litres ✬✬ Distance = 225 KM ✬Step-by-step explanation:Given:A car consumes 10 litres of petrol to cover a distance of 150 km.To FIND:How much petrol will it consume for covering distance of 450 ?How much distance will it cover in 15 litres of petrol ?Solution: LET's find out , how much petrol the car will consume for covering 1 km.[ For this divide the distance, 150 km by 10 litres ]➟ 150/10 ➟ 15 kmSo the car consumes 1 litre of petrol for covering distance of 15 km.Now let , for covering a distance of 450 the car will consume X litres. x = Distance/15 x = 450/15 x = 30 litresHence, the car will consume 30 litres of petrol for covering distance of 450 km.________________________Now , let car will cover a distance of x km in 15 litres of petrol.x = 15 × distance covered in 1 litre x = 15 × 15 x = 225 kmHence, the car will cover a distance of 225 km in 15 litres of petrol.

8.3Step-by-step EXPLANATION:

8.6667 is ANSWER. OKK FRIEND

1 RUPEE coin = 10 ml 8 rupee COINS = 8 × 10 ml = 80 ml ,,,,,is answer of this

so MUCH 100% RIGHT ANSWER

5600 & 5000hope it helps!!!!!PLZ MARK me as BRAINLIEST!!!!!

Answer:0.0006026Answer:0.0006026Step-by-step explanation:3.126/5187.4=3126×10/5187×10003.126/5187.4=3126×10/5187×1000=3126/51874×1003.126/5187.4=3126×10/5187×1000=3126/51874×100=1563/25937×1003.126/5187.4=3126×10/5187×1000=3126/51874×100=1563/25937×100=1563/25937003.126/5187.4=3126×10/5187×1000=3126/51874×100=1563/25937×100=1563/2593700=0.00060263.126/5187.4=3126×10/5187×1000=3126/51874×100=1563/25937×100=1563/2593700=0.0006026pls MARK me as a BRAINLIEST

56usg+×[EMAIL PROTECTED]+272

1320 RsStep-by-step explanation:Since the GARDENER MAKES 2 rounds of the fenceThus,LENGTH of rope required=2×circumfrece of the CIRCULAR gardendiameter=42mradius=diameter/2radius=42/2=21mCircumference of the circular grden=2πr=2×22/7×21=2×22×3=44×3=132length of rope required=2×circumference of the circular garden=2*132=164 mcost of rope for 1 m=5 Rscost of 264 m=264×5=1320 Rs

-STEP EXPLANATION:the above is your answer so MARK me as a brainlist frnd

N : ❒ Given :- Rate of INTEREST, R = 10% p.a.No. of years, n = 3 years Difference between C.I. and S.I. = Rs. 496❒ To Find :- The sum of money, P = ? ❒ Formulas Used :- ❒ Calculation of Simple Interest :- ❒ Calculation of Compound Interest :- According to Question,Hence, the sum of money, P = Rs. 16000.VERIFICATION : ❒ We have :- Sum of money, P = Rs. 16000Rate of Interest, r = 10% No. of years, n = 3 years. ❒ To VERIFY :- Difference between C.I. and S.I. = Rs. 496Now, And,Hence, Verified.ANSWER : ❖ If the difference between the compound interest and the simple interest on a CERTAIN sum of money at 10% per annum for 3 years is Rs. 496, then when the interest is compounded annually the sum is Rs. 16000.

-STEP EXPLANATION:HOPE this is CORRECT ANSWER

sghgfgfdfffbgfffddd I am doing

x=8, y=3Step-by-step explanation: 22/x+y + 15/x-y = 5         [1]55/x+y  + 40/x-y = 13       [2]LCM of 40 and 15 is 120Therefore, multiplying [1] by 8 and [2] by 3, we get[1] 8 x (22/x+y + 15/x-y =5]-------> 176/x+y + 120/x-y = 140 [3][2] 3 x (55/x+y + 40/x-y =13]------>165/x+y + 120/x+y = 139 [4]Now, Subtracting [4] from [3],  we get.   176/x+y + 120/x-y  = 140   165/x+y + 120/x-y  = 139   -             -                   -   [Changing the signs, as we're subtracting]____________________11/x+y                         = 1 ----------------> 11=x + y [5]now, substituting the value of (x+y) in eq. 2we get,22/11 + 15/x-y =5---------> 2 + 15/x-y =5--------> 15/x-y =3--------> 15=3x-3yOn dividing the whole eq. of 15=3x-3y by 3, we get 5=x-y[6]now ADDING [5] and [6]11=x+y5=x-y______      [no CHANGE in signs, as we're adding]16=2x------> x=8substituting the value of 'x' in eq. 6, we get5=8-y-----> 5-8 = -y-----> -3 = -y ------> 3=yHope this answers your query.

From the QUESTION, DEALER SELLING a sofa set for =Rs.21600= Selling price He gain on selling =8% COST price of sofa set, CP=Rs.{(100/(100+gain%))×SP} ={(100/(100+8))×21600} ={(100/108)×21600} ={2160000/108} =Rs.20000

-STEP EXPLANATION:let the number be xlet the number be x9:15=15:xlet the number be x9:15=15:XX=(15*15)/9let the number be x9:15=15:xx=(15*15)/9x=25

Hence, the NUMBER is 72. This DISCUSSION on A number is 56 greater than the AVERAGE of its third, QUARTER and one-twelfth.

I don't KNOWN SORRY bruhhhhhhhhhhhhhhhhhhhh

It is TRUE that it was CONSTRUCTED by RED STONE

I UC BOY X SUSSEX HADDOCK

-STEP explanation:can you PLEASE SEND photo of other sides of TRIANGLE

20000Step-by-step EXPLANATION:From the question,Dealer SELLING a sofa SET for =Rs.21600= Selling priceHe gain on selling =8%COST price of sofa set,CP=Rs.{(100/(100+gain%))×SP}={(100/(100+8))×21600}={(100/108)×21600}={2160000/108}=Rs.20000

5x=145x= 145/5x= 29Step-by-step EXPLANATION:HOPE it will HELP you

.12 -3.92 = 7.82-3.92= 3.9

(ii) cricket is played maximum NUMBER of students(I) It's GIVING GAMES information

-step EXPLANATION:the above is your answer so MARK me as a BRAINLIST frnd and

34.86is the CORRECT ANSWER

-STEP EXPLANATION:what is this? I don't KNOW this THING what is this?

B = BracketO = of D = DivisionM = MultipicationA = AdditionS = SubractionStep-by-step EXPLANATION:It's an order of doing problem in expressions and EQUATIONS. MARK as Brainliest

डिफरेंशियल ईक्वेशंस) उन संबंधों को कहते हैं जिनमें स्वतंत्र चर तथा अज्ञात परतंत्र चर के साथ-साथ उस परतंत्र चर के एक या अधिक अवकल गुणांक (डिफ़रेंशियल कोइफ़िशेंट्स) हों।{\displaystyle F\left(X,y,Dy,D^{2}y,\ldots ,D^{n}y\right)=0}{\displaystyle F\left(x,y,Dy,D^{2}y,\ldots ,D^{n}y\right)=0} एक अवकल समीकरण का सामान्य रूप है।यदि इसमें एक परतंत्र चर तथा एक ही स्वतंत्र चर भी हो तो संबंध को साधारण (ऑर्डिनरी) अवकल समीकरण कहते हैं। जब परतंत्र चल तो एक परंतु स्वतंत्र चर अनेक हों तो परतंत्र चर के खंडावकल गुणक (PARTIAL DIFFERENTIALS) होते हैं। जब ये उपस्थित रहते हैं तब संबंध को आंशिक (पार्शियल) अवकल समीकरण कहते हैं। परतंत्र चर को स्वतंत्र चर के पर्दो में व्यंजित करने को अवकल समीकरण का हल करना कहा जाता है।यदि अवकल समीकरण में nवीं कक्षा (ऑर्डर) का अवकल गुणक हो और अधिक का नहीं, तो अवकल समीकरण nवीं कक्षा का कहलाता है। उच्चतम कक्षा के अवकल गुणक का घात (पॉवर) ही अवकल समीकरण का घात कहलाता है। घात ज्ञात करने के पहले समीकरण को भिन्न तथा करणी चिंहों से इस प्रकार मुक्त कर लेना चाहिए कि उसमें अवकल गुणकों पर कोई भिन्नात्मक घात न हो।अवकल समीकरण का अनुकलन सरल नहीं है। अभी तक प्रथम कक्षा के वे अवकल समीकरण भी पूर्ण रूप से हल नहीं हो पाए हैं। कुछ अवस्थाओं में अनुकलन संभव हैं, जिनका ज्ञान इस विषय की भिन्न-भिन्न पुस्तकों से प्राप्त हो सकता है। अनुकलन करने की विधियाँ सांकेतिक रूप में यहाँ दी जाती हैं।प्रयुक्त गणित, भौतिक विज्ञान तथा विज्ञान की अन्य शाखाओं में भौतिक राशियों को समय, स्थान, ताप इत्यादि स्वतंत्र चलों के फलनों में तुरंत प्रकट करना प्राय: कठिन हो जाता है। परंतु हम उनकी वृद्धि की दर तथा उसके अवकल गुणकों में कोई संबंध बहुधा बड़ी सुगमता से पा सकते हैं। इस प्रकार ऐसे अवकल समीकरण प्राप्त होते हैं जिन्हें पूर्वोक्त राशियाँ संतुष्ट करती हैं। इन्हें हल करना उन राशियों का ज्ञान प्राप्त करने के लिए आवश्यक होता है। इसलिए विज्ञान की उन्नति बहुत अंश तक अवकल समीकरण की प्रगति पर निर्भर है।

s only ONE number between 61 and 107 that is a multiple of 3, 5 and 9.That number is 90.An easy way to figure this out would be to start with the 9 table…starting in the 60’s.If it is DIVISIBLE by, it will also be a multiple of 3. So we really just NEED a number that is a multiple of both 9 and 5.…63 72 81 90 ahhh 90 is divisible by 3, 5 and 9 :)

-STEP EXPLANATION:GIVEN:-sinA=1/√2,tanB=1.To PROVE:-If sinA=1/√2,tanB=1 then prove that cos(A+B)=0Solution:-Given that-SinA=1/√2=>SinA=Sin 45°A=45°------(1)And TanB=1=>TanB=Tan 45°B=45°--------(2)Now LHS:-Cos(A+B)=>Cos(45°+45°)=>Cos90°=>0=RHSLHS=RHSHence Proved.

-step explanation:⟹numberofchairs=85\tt \blue{⟹ \: cost \: of \: 1 \: CHAIR \: is \: rupees180}⟹costof1chairisrupees180\tt \orange{⟹ \: cost \: of \: 85 \: chairs = 85 \times 180}⟹costof85chairs=85×180_____________________________________Given:\tt \pink{⟹ \: number \: of \: TABLES = 25}⟹numberoftables=25\tt \red{⟹ \: cost \: of \: one \: tables \: is \: RUPEES \: 140}⟹costofonetablesisrupees140\tt \blue{⟹ \: cost \: of \: 25 \: tables = 25 \times 140}⟹costof25tables=25×140__________________________________Now.total cost of chairs and tables = (85×180 + 25× 140)\rm{⟹(85 \times 180 + 25 \times 140}⟹(85×180+25×140\rm{⟹ \: = 15300 + 3500}⟹=15300+3500\rm{⟹ \: = 18800}⟹=18800________________________________Now,Money GIVEN in advance is 2500balance money to be paid to the dealer is\tt{⟹ \: = 18800 - 2500}⟹=18800−2500\tt \red{ = 16300}=16300please Mark me as BRAINLIEST answer

-STEP EXPLANATION:vxhfzxyfxjg B JV ufx6rz6ez6dxydxydx6rz6r5dx6rxt