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  • 1 answers

Avatar ? 5 years, 1 month ago

All books are good dear....but, you just have to do hard work..?
  • 2 answers

Yogita Ingle 5 years, 1 month ago

Let a be any positive integer and b = 4. Then by Euclid’s algorithm,
a = 4q + r for some integer q ≥ 0, and r = 0, 1, 2, 3
So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3 because 0 ≤ r < 4
Now, 4q i.e., 2(2q) is an even number
∴4q + 1 is an odd number.
4q + 2 i.e., 2(2q + 1) which is also an even number.
∴ (4q + 2) + 1 = 4q + 3 is an odd number.
Thus, we can say that any odd integer can be written in the form 4q + 1 or 4q + 3 where q is some integer.

Shreedhar J 5 years, 1 month ago

By Euclid division lemma, a=4q+r r=0,1,2,3 Case(i).r=0, a=4q=2(2q)=2m[even] (ii). r=1,a=4q+1=2(2q)+1=2m+1[even]
  • 1 answers

Yogita Ingle 5 years, 1 month ago

1445 = 1190*1 + 255
1190 = 255*4 + 170
255 = 170*1 + 85
170 = 85*2 + 0
So, now the remainder is 0, then HCF is 85
Now,
85 = 255 - 170
(1445 - 1190) - (1190 - 255*4)
⇒ 1445 - 1190 - 1190 + 255*4
⇒ 1445 - 1190*2 + (1445 - 1190)*4
⇒ 1445 - 1190*2 + 1445*4 - 1190*4
⇒ 1445*5 - 1190*6
⇒ 1190*(- 6) + 1445*5
1190m + 1445n , where m = - 6 and n = 5

  • 4 answers

Muskan Shekhawat 5 years, 1 month ago

896

Geeta Jadav 5 years, 1 month ago

2(112)4=224×4=896

Raghav Jangir 5 years, 1 month ago

Or koi questions

Raghav Jangir 5 years, 1 month ago

8(56+56) =448+448 =896
  • 1 answers

Saloni Pathak 5 years, 1 month ago

1170=650 multiply 1+520 650=520 multiply 1+130 520=130multiply 4+0 130 is the answer
  • 1 answers

Kashish . 5 years, 1 month ago

5. Is it possible to design a rectangular park of perimeter 80 metres and area 400 m2. If so, find its length and breadth. Ans. Let length of park = x metres We are given area of rectangular park =  Therefore, breadth of park =  metres{Area of rectangle = length × breadth} Perimeter of rectangular park = 2 (length + breath) = metres We are given perimeter of rectangle = 80 metres According to condition: =>  ⇒  ⇒  ⇒  ⇒  Comparing equation,  with general quadratic equation , we get a = 1, b = −40 and c = 400 Discriminant =  (1) (400) = 1600 – 1600 = 0 Discriminant is equal to 0. Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter 80 metres and area . Using quadratic formula  to solve equation,  Here, both the roots are equal to 20. Therefore, length of rectangular park = 20 metres Breadth of rectangular park = 
  • 1 answers

Gaurav Seth 5 years, 1 month ago

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:

(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Solution:

Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.


(i) From the given condition,

ΔABC ~ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Since, ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)


In ΔACD and ΔFGH,

∠A = ∠F

∠ACD = ∠FGH

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)

∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

∴ ΔDCA ~ ΔHGF (AA similarity criterion)

  • 2 answers

Yangzee Sherpa 5 years, 1 month ago

Let us consider that 3√2 is a rational number. It can be written in the form p/q (p and q are co-primes). →p/q = 3√2. →p/3q = √2. Now, p/3q = integer/integer = rational number. But, this contradicts the fact that √2 is irrational. Therefore, our assumption that 3√2 is rational is WRONG. Hence, 3√2 is an irrational number. HoPe It HeLpS yOu??

Gaurav Seth 5 years, 1 month ago

Let us consider that 3root2 is a rational number. It can be written in the form p/q (p and q are co primes)

p/q = 3root2

p/3q = root2

Now,

p/3q = integer/interger

= rational number

But, this contradicts the fact that root2 is irrational.

Therefore, our assumption that 3root2 is rational is WRONG.

Hence, 3root2 is an irrational number.

  • 3 answers

Mortal Combat 5 years, 1 month ago

0.04 or 0

Mortal Combat 5 years, 1 month ago

hope it will help you

Mortal Combat 5 years, 1 month ago

0.04and 0
  • 1 answers

Abhinav Chahar 5 years, 1 month ago

a=-10 Common difference=4
  • 1 answers

Gaurav Seth 5 years, 1 month ago

If Tan A = 3/4, then find the other trigonometric ratio of angle A.

sol) Tan A = Opposite side = 3
                      Adjacent side    4

.^., opposite side : adjacent side = 3:4

For angle A, 

opposite side = BC=3k

Adjacent side =AB=4k(where k is any +ve no)

Now, we have in triangle ABC( by Pythagoras theorem) 
 



AC^2 = AB^2 + BC^2

=>(3k)^2 + (4k)^2 = 25k^2

AC = √25k^2

=> 5k = Hypotenuse 

Now,we can easily write the other ratios of trigonometry

sin A = 3k 3
             5k    5

cos A = 4k 4
             5k    5

cosec A =   1      =  5
                 sin A     3


sec A =   1     =   5
             cos A     4


cot A =   1      =  4
             tan A     3

  • 1 answers

Gaurav Seth 5 years, 1 month ago

 

Let the speed of car at A be x kmph

and the speed of car at B be y kmph

As per the question,

5x - 5y = 100

 x - y = 20 ... (1)

and x + y = 100 ... (2)

Solving (1) and (2), we get,

x = 60 and y = 40

 Speed of the car at A = 60 kmph

Speed of the car at B = 40 kmph

  • 3 answers

Ritika Ritika 4 years, 10 months ago

Thanks

Ritika Ritika 5 years, 1 month ago

And please tell their contribution

Prabhat Mishra 5 years, 1 month ago

Aryabhata. ... Brahmagupta. ... Srinivasa Ramanujan. ... P.C. Mahalanobis. ... C.R. Rao. ... D. R. ... Harish Chandra. ...
  • 1 answers

Gaurav Seth 5 years, 1 month ago

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1 
=2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1
=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2
-2sin2θcos2θ]+1
The algebraic identity
a3 + b3 = (a+b)3 - 3ab(a+b) and 
a+ b= (a+b)2 - 2ab
are used in the above step where
a = sin2θ and b = cos2θ.
writing sin2θ + cos2θ = 1, we have 
= 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1
= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1
= -3+3=0                 

  • 2 answers

Priyanshu Kumar 5 years, 1 month ago

So breadth =16 & length =33

Priyanshu Kumar 5 years, 1 month ago

Let the breadth be x Therefore length = 2x+1 (2x+1)x= 2x² +x= 528 m² {Since area of rect■ is l×b} 2x²+x= 528 2x²+x-528=0 2x²-32x+33x-528=0 2x(x-16)+33(x-16)=0 (x-16)(2x+33)=0 Therefore, x=16 Neglecting x=-33/2 {as a side can't be negative}
  • 1 answers

Himanshu Nagaich 5 years, 1 month ago

Let A(1,-1),B(5,2) and C(9,5) Area of Triangle=1/2[X1(Y2-Y3)+X2(Y3-Y1)+X3(Y1-Y2)] =1/2[1(2-5)+5(5+1)+9(-1-3)] =1/2[-3+30-36] =1/2*(-9) =4.5 unit sq.(AREA CANNOT BE NEGATIVE)
  • 2 answers

Harini Reddy 5 years, 1 month ago

answer is 1

Gaurav Seth 5 years, 1 month ago

 

A n s w e r :

=> (sin30+cos30) - (sin60+cos60)

 

sin30 = 1/2

 

sin60= √3/2

 

cos30= √3/2

 

cos60= 1/2

 

=> (1/2+√3/2) - (√3/2+1/2)

 

=> 1+√3/2 - √3+1/2

 

=> 1+1+√3-√3 /2

 

=> 2/2

 

=>1

  • 2 answers

Harini Reddy 5 years, 1 month ago

how can we consider 8 as a middle term can please give step by step explanation

Tanu Man 5 years, 1 month ago

We will consider 8 as a middle term
  • 1 answers

Geeta Jadav 5 years, 1 month ago

Let length of rectangular park = x Breadth of rectangular park=y Now, Area of rectangal=l×b 400=x×y.......(1) Perimeter of rectangal=2(l+b) 80=2(x+y) 40=x+y X=40-y.........(2) Now, 400=xy 400=(40-y)y 400=40y-y^2 -y^2+40y-400=0 y^2-40y+400=0 y^2-20y-20y+400=0 y(y-20)-20(y-20)=0 (y-20)(y-20)=0 y=20m Now,put the value of y in eq(2), X=40-20 x=2m Hence, value of length =20m Value of breadth=20m
  • 5 answers

Tulsi Yadav 5 years, 1 month ago

110

Harini Reddy 5 years, 1 month ago

110

Chetan Soni 5 years, 1 month ago

110

........ ...... 5 years, 1 month ago

110

Seema Choudhary 5 years, 1 month ago

110

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