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Riley Joseph 5 years ago

Suppose you have to find out 29% of something (x) so we will take that value 29/100*x and see the result And if u were asking in regard to gst then 29/100*x+x =amt of bill
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Gaurav Seth 5 years, 3 months ago

Answer:

first let the points be

A(4,-1),B(-2,-3)

first let us take the ratio as 1:2 and later 2:1 since its trisection  .

let the point joining A and B be P(x,y)

Step-by-step explanation:

P = (1*-2+2*4/1+2  ,  1*-3+2*-1/1+2)

  = (-2+8/3  ,  -3-2/3)

  =  (6/3  ,  -5/3)

P   =  (2  ,  -5/3)

now let ratio be 2:1

P = (2*-2+1*4/3  ,  2*-3+1*-1/3)

  = (-4+4/3  ,  -6-1/3)

  = (0/3  ,  -7/3)

P = (0 , -7/3)

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Riley Joseph 5 years, 2 months ago

Inverse means just change the numeretor and denominator Example: 1).4/7=7/4

Ram Kushwah 5 years, 3 months ago

In verse mein numerator denominator ho jata hai aur

denominator numerator ban jata hai

example 

3/4 ka inverse =4/3

7=7/1 ka invese=1/7

1/9 ka invese=9/1=9

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Saivaling Appa 5 years, 4 months ago

€£2
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Ranja Naskar Halder 5 years, 4 months ago

1
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Sia ? 5 years, 4 months ago

p = ab2 = a x b x b
q = a3b = a x a x a x b
LCM(P, Q) = a b
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Ram Kushwah 5 years, 3 months ago

secθ= cosec60

=cosec(90-30)=sec30 ( As cosec(90-θ)=secθ)

so θ=30

Now 2cos2θ-1

=2cos60-1

=2*1/2-1

=1-1=0

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Lokesh C S 5 years, 5 months ago

a + 2d = 14 a + 3d = 18 On solving these equations - d = -4 or d= 4 and a = 6. AP will be 6 , 14 , 18 , ........ n= 51 , a = 6 and d = 4 S 51 = n/2[2a + (n-1) d] Substituting the values , S51 = 51/2 [ 2×6 + ( 51 - 1 ) 4 ] = 51/2 [ 12 + 200 ] = 51/2 × 212 = 51 × 106 = 5406
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Gaurav Seth 5 years, 6 months ago

Explanation:

S.P of Rs. 5000 stock = Rs.(156100*5000)Rs.156100*5000= Rs. 7800. 

 

Income from this stock = Rs.(12100*5000)Rs.12100*5000 = Rs. 600.  

 

Let investment in 8 % stock be x and that in 9 % stock = (7800 - x).

 

Therefore,  

 

(x*890)+(7800−x)*9108=[600+70]x*890+7800-x*9108=600+70  

 

4x45+7800−x12=670⇔x=36004x45+7800-x12=670⇔x=3600 

 

Therefore,  Money invested in 8 % stock at 90 = Rs. 3600.

 

Money invested in 9 % at 108 = Rs. (7800-3600) = Rs. 4200.

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Unish Thapa 5 years, 6 months ago

1
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Yogita Ingle 5 years, 7 months ago

Let us use a  trick to solve it quickly

99x + 101y = 499 ............ i

101x + 99y = 501............ ii

subtract i  from  ii

2x - 2y = 2
x - y =1
x = 1 + y   ........   iii

add i and  ii

200x + 200y = 1000
x + y = 5 .................................. iv

Now from iii and  iv  ,  we get  1 + y + y = 5       or          2y = 4       or        y = 2     ......  v

again from iii and  v   , we get  x = 1 + 2 = 3

Hence x = 3  ,  y = 2

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Yogita Ingle 5 years, 7 months ago

5x + 2y = -3 ........ (i)
x + 5y = 4
x = 4 - 5y ........... (ii)
Put (ii) in (i), we get
5(4 - 5y) + 2y = -3
20 - 15y + 2y =-3
- 13y = - 3 - 20
-13y = -23
y = 23/13
Put value of  y in (ii), we get
x = 4 - 5(23/13)
x = (52 - 5)/13 = 47/3
 

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Soma Kar Dutta 5 years, 6 months ago

Each child got 8 books. As ( 32/4=8) In the question Fatima is also considered as a child.
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Yogita Ingle 5 years, 7 months ago

5m - 3n = 19 ......... (i)
m - 6n = -7 ........... (ii)
Multiple (i) by 2 , we get
10m - 6n = 38........... (iii)
Subtract (iii) and (ii), we get
(10m - 6n)- (m - 6n) = 38 - (-7)
10m - 6n - m + 6n = 38 + 7
9m = 45
m = 5
Put m = 5 in (ii) we get
5 - 6n = -7
-6n = -7 - 5
-6n = -12
n = 2

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Yogita Ingle 5 years, 7 months ago

3x + 2y = 29  .......... (i)
5x - y = 18  .......... (ii) 
multiply equation (ii) by 2 and add in (i) ,  we get
3x + 2y = 29
+10x - 2y = 36
____________
13x=65
x = 65/13=  5
now put in (ii), we get
5(5) - y = 18
25 - y = 18
25 - 18 = y
7 = y
and x = 5


 

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Kunal Nawariya 5 years, 3 months ago

1/x -1/x-2=3 Take Lcm X-2-X /x(x-2)=3 -2/x^2-2x =3 -2=3(x^2-2x) -2=3x^2-6x 0=3x^2-6x+2 Hera,a=3,b=-6,c=2 D=b^2-4ac=(-6)^2-4*3*2=36-24=12 Since D>0 So x= -b+-√D/2a= 6+-√12/2*3= 6+-√12/6=6+√12/6,6-√12/6
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Lokesh C S 5 years, 5 months ago

tan A + cot A = 2 , to prove tan^2 A + cot^2 A = 2 Squring (tan A + cot A )^2 = tan^2 A + cot ^2 A + 2 × tan A × cot A = tan^2 A + cot^2 A + 2 × tan A × 1/tan A ( cot A is inverse of tan A ) = tan^2 A + cot^2 A + 2 (2)^2 = tan^2 A + cot^2 A + 2 4 - 2 = tan ^2 A + cot^2 A 2 = tan^2 A + cot^2 A hence provef

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