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this is wrong right answer is 2a -ba

Sp=Sq

p/2[2a +(p-1)d]=q/2 [2a+(q-1)d]

p[2a +pd - d]=q[2a+qd - d ]

2ap + p^2d - pd =2aq + q^2d -qd

2ap-2aq +p^2d -q^2d -pd +qd =0

2a (p-q) +(p+q)(p-q)d -d(p-q)=0

(p-q)[2a + (p+q)d - d ]=0

2a + (p+q)d - d=0

2a + [(p+q)-1]d=0

QR = 1.5 = (asset - stock)/liabilities

=> (100000-stock) = 1.5*(40,000)=> stock = 100000-60000 = 40000

food=500*70 people days.

food=700*days

same food, so

days=500/700 * 70=50 days.

thank u

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Given, sum of first P termsSp = ap^2 + bp.

Then sum of first two terms.S2 = a(2*2) + b*2 = 4a + 2b

First term of an APt1 = a(1*1) + b(1)4 3 = a + b

Second term of an APt2 = (4a + 2b) - (a + b) = 3a + b

Common difference= t2 - t1= (3a + b) - (a + b)= 2a

3192 + ? + 2112= 65145304 + ? =6514? = 6514-5304? = 1210

3192+?+2112=6514?=1210

4x -12>-0; x = 3, 4(3)=12;

4x >_12;. x>_12/4=3

tan^-1(tanx) = x

So tan^-1(tan 3π/4) = 3π/4

Like my answer if you find it useful!

The given inequality is 2x-1≤x²+3≤x-1 => 2x-1-3 ≤x²≤x-1-3=> 2x-4 ≤ x²≤ x-4

since 2x-4 will be > 0 oy when x > 2 , and at x > 2 , x² will always be greater tha 0.. so ,the first inequality holds true ✓

for second , x² will never be greater than x-4 because x-4 wil be > 0 when x > 4 , and then the value of x² will always be greater than 0

so, it will not hold true..

hence no solution is there.. for this inequality.

wrong answer

Distance travelled = 8700-6514

= 2186 km

Let the dist. be xdist. while going = dist while returningtime = dist./speedso, time while going = x/20time while coming =x/30total time= x/20+x/30=(3x+2x)/60=5x/60=x/12total dist.= 2xavg. speed = total dist./total speed=2x*12/x=24km/hr

given , speed of car =20,30km/hraverage speed =totaldistance/time takentime =2hr because 20km per hour ,30 km per hour both are hoursboth time =1+1=2average speed =20+30/2=50/2=25km/hr

Let the distance traveled by Abdul from home to school = s kmtime taken to reach the school=t1 secFor Return journey, Abdul cover distance =s kmtime=t2 sec∴ Average speed for forward journey[home - school ] = Total distance/ Total time.20 km/h =s/t1∴t1=s/20 h ------eq(1)Average speed for backward journey[school -home] = Total distance/ Total time.30 km/h =s/t2t2=s/30h -----eq(2)Average distance for entire journey = Total distance/ Total time

=(s+s)/[s/20 +s/30] =2s/s[1/20+1/30] =2x20x30/50 =24 km/hrShortcut method : If equal distance are covered with speeds v1 and v2 then average distance=2v1v2/v1+v2 = 2x20x30/50 =24km/hr

∴ The average speed for Abdul's trip is 24km/h

cos 56+ cot 56= cos (90-34)+ cot (90-34)= sin 34+ tan 34

cos(90-56)+cot(90-56)=sin 34+tan 34

thought I. Hugh hub hanging on I it was back in your way

cos56+ cot56= cos(90-34)+cot(90-34)=sin34+ tan34

The ans will

= sin34 + tan34

sec 34°=x=>sec² 34°=x²=>1+tan² 34°=x²=>tan² 34°=x²-1=>tan²(90°-56°)=x²-1=>cot² 56°=x²-1

And sec 34°=x=>sec(90°-56°)=x=>cosec 56°=x

∴cot² 56°+cosec 56°=x²+x-1

basic addition and subtraction56-48+56-23-465+4988+55-5000=-381

pr is parrallel to S line by linear pair their sum is 180

x=125°because they were same angles

125 because corresponding angles are equal

x=125because they corresponding angles

x=125Because,corresponding angles are equalso,ans 125

987 = 3 × 7 × 47

10500 = 2 × 2 × 3 × 5 × 5 × 5 × 7

HCF of 987 and 10500 = 3 × 7 = 21

Therefore ,

Least form of the fraction = 987 / 10500

Divide numerator and denominator with HCF,

we get,

= ( 987 / 21 ) / ( 10500 / 21 )

= 47 / 500

987 / 10500 = 47 / 500

Denominator = 500

= 2^2 × 5^2

We have only 2 and 5 as factors .

Therefore ,

987 / 10500 is terminating decimal.

thanks

1350 cube root

112 one hundred and twelve

56+56=112or2×56=112

the answer is 112feet

112 is your required answer......

112 one hundred and twelve

112 is the right answer

From the above the diagrams CM and RNare the medians for the triangles.

we know thatΔABC≈ΔPQR

AB/PQ = BC/QR = AC/PR ------------------------(1)fromΔAMC andΔPNR∠A =∠P∠M =∠N = 90AC = PRby ASA similarityΔAMC≈ΔPNR ----PROVED

from thisAC/PR = CM/RN = AM/PNFROM ------(1)AB/PQ = BC/QR = AC/PRfrom thatCM/RN = AB/PQ -----PROVED

fromΔBMC andΔQNR∠B =∠Q∠M =∠N = 90BC = RQby ASA similarityΔBMC≈ΔQNR ------PROVED

plz fast

thank u

The theory of arithmetic says that a no needs to have only 2 and 5 as it's prime factors in order to be a terminating one

so here 17/2*3^2*5.

Probability of getting a head in toss of coin = 1/2

Probability of getting 4 in a throw of dice = 1/6

Therefore, chances of getting head in toss of coin is more than getting 4 in a throw of dice.

4x + 74 correct

(5^2×t^(-4))/(5^(-3)*5*2*t^(-8))5^(2+3-1)*t^(-4+8)/25^4*t^4/2{(5t)^4}/2

1/(5*6) + 1/(6*7) + 1/(7*8) + 1/(8*9) + 1/(9*10) + 1/(10*11) + 1/(11*12)= (1/5 - 1/6) + (1/6 - 1/7)+ (1/7 - 1/8) + (1/8 - 1/9) + (1/9 - 1/10) +(1/10 - 1/11) + (1/11 - 1/12)= 1/5 - 1/12= 7/60

6/55

answer malum ho to bta dena hmko v

Griffin always wanted to be invisible, but as soon as his wish was fulfilled, he started to face the consequences.

In order to be invisible, he had to remove his clothes and it was impossible to survive in the cold of London without clothes. The next disadvantage of being invisible was he couldn’t eat anything in public. He didn’t want to confide himself and hence he couldn’t take shelter in anybody’s home.

Although he was invisible, dogs could smell him and followed him wherever he went.

Therefore, Griffin wanted to be visible again but couldn’t get a quiet place to perform his experiments.

thnks

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