-Answered by Master Purushottam On 22 August 2019:09:17:36 PM

.

-Answered by Master Purushottam On 22 August 2019:09:18:51 PM

 Aryabhatta discovered zero.

-Answered by Akki chauhan On 19 August 2019:09:50:45 AM

 X-2=7

 X=7+2

 X=9

-Answered by Akki chauhan On 18 August 2019:08:35:07 PM

 Let the quadratic polynomial be ax²+bx+c=0 and its zeroes be α and β.

sum of zeroes     α+β = -3 = -b/a 

product of zeroes      αβ = 2= c/a

If a=1   then b=3,  c=2

So, the quadratic polynomial is x²+3x+2.

-Answered by Himanshi Verma On 18 August 2019:11:00:35 AM

 By Thales theorem or basic propositionality Theorem

-Answered by Himanshi Verma On 18 August 2019:10:58:35 AM

 Mean, Mode, Median are the measurment of central tendency.  mean and mode cann't be shown on graph but median can be shown on the graph.

-Answered by Himanshi Verma On 17 August 2019:10:34:55 AM

 The general form of an Arithmetic progression is a, a+1, a+2, a+3.....

-Answered by Himanshi Verma On 17 August 2019:10:41:29 AM

-Answered by Himanshi Verma On 18 August 2019:10:41:56 AM

-Answered by Himanshi Verma On 18 August 2019:11:09:55 AM

-Answered by Himanshi Verma On 23 July 2019:07:52:25 PM

-Answered by Himanshi Verma On 18 July 2019:09:05:33 PM

Given: a₁ = 6, S₆ = 66 

using Formula: S_n = n2 (a + l) 

Step by Step Explanation: 

S₆ = 62 ( 6 + a₆)

66 = 3(6 + 6^th term) 

(6 + 6^th term) = 663 = 22

6^th term = 22 - 6 

6^th term = 16 Answer 

-Answered by Master Mind On 11 July 2019:09:08:06 PM

Given equation is 6(7b - 4) = 60  Step by Step explanation

7b - 4 = 606 

7b - 4 = 10 

7b = 10 + 4 

7b = 14 

b = 147 = 2 Answer

-Answered by ATP Admin On 09 July 2019:11:53:03 PM

Given:S_n = 4n² - 7n 

Step by step explanation: 

(i) Sum of 23 terms 

S_n = 4n² - 7n ............. (i) 

Replace n by 23 we have,

S₂₃ = 4(23)² - 7(23) 

= 4 × 529 - 161

= 2116 - 161 

= 1955 Answer

(ii) 15^th term 

a₁₅ = S₁₅ - S₁₄ 

S₁₅ = 4(15)² - 7(15) = 4 × 225 - 105 = 900 - 105 = 795 

S₁₄ = 4(14)² - 7(14) = 4 × 196 - 98 = 784 - 98 = 686

Now, a₁₅ = S₁₅ - S₁₄ 

              = 795 - 686 

              = 109 Answer

(iii) n^th term 

a_n = S_n - S_n-1

Given, S_n = 4n² - 7n 

Replace n by n-1 we have 

S_n-1 = 4(n-1)² - 7(n-1) 

       = 4(n² - 2n + 1) - 7n + 7

       = 4n² - 8n + 4 - 7n + 7 

       = 4n² - 15n + 11 ............. (ii) 

Now Applying equation (i) and (ii) in formula a_n = S_n - S_n-1 

a_n = (4n² - 7n) - (4n² - 15n + 11) 

    = 4n² - 7n - 4n² + 15n - 11

    = 8n - 11 

Therefore, a_n = 8n - 11 Answer  

*******************************

-Answered by Master Mind On 11 July 2019:09:42:19 PM

 x²+ 3x+7 

This equation is of the form of ax²+bx+c=0

so, a=1,     b=3,    c=7

b²+4ac = 3×3 - 4×1×5

           = 9 - 28 

            = 19 

∴  19 > 0

So, the given equation has two real roots.

-Answered by Himanshi Verma On 18 July 2019:07:35:08 PM

-Answered by Master Purushottam On 08 July 2019:10:58:57 PM

-Answered by ATP Admin On 08 July 2019:10:45:29 PM

 This is an AP a=6,d=6, and n=40

Using formula,  

S_n = n2 [2a+(n-1)d]

S₄₀= 402 [2*6+(40-1)6]

S₄₀=20(12+234)

S₄₀=20*246

S₄₀=4920

-Answered by Akki chauhan On 18 July 2019:10:26:06 AM

Solution: Using Euclid's division lemma any positive integer can be written in the form of a = bq + r where r = 0, 1, 2 ...... and q is quotients. 

Let the number which is divisible by 3 be 3q + 0 or 3q + 1 or 3q + 2 where [0 <= r < b] 

Now n = 3q or n = 3q + 1 or n = 3q + 2 

Case I, 

When n = 3q  .......... (i) 

⇒ n = 3(q) where n is divisible by 3

Adding 2 both sides in equ. (i) 

We have,

    n + 2 = 3q + 2 Where n + 2 is not divisible by 3 

Now adding 4 both side in equ. (i) 

We have,

    n + 4 = 3q + 4 Where n + 4 is not divisible by 3 

Case II  

  Taking n = 3q + 1 ........ (ii) where n is not divisible by 3

Adding 2 both sides in equ. (ii) 

We have, 

n + 2 = 3q + 1 + 2 = 3q + 3 

n + 2 = 3(q + 1) where n + 2 is divisible by 3 

Now adding 4 both sides in equ. (ii) 

n + 4 = 3q + 1 + 4 = 3q + 5 where n + 4 is not divisible by 3 

Case III  

taking n = 3q + 2 .... (iii) where n is not divisible by 3 

Adding 2 both sides in equ. (iii) 

We have,

n + 2 = 3q + 2 + 2 = 3q + 4 where n + 2 is not divisible by 3

Now adding 4 both sides in equ. (iii) 

We have, 

n + 4 = 3q + 2 + 4 = 3q +6 

n + 4 = 3(q + 2) where n + 4 is divisible by 3 

Hence in all these three cases we have seen that either one and only one n or n + 2 or n + 4 is divisible by 3. 

-Answered by ATP Admin On 17 June 2019:11:28:43 PM

 The rational numbers between 7 and 9 are 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 90. 

-Answered by Himanshi Verma On 14 May 2019:11:11:27 AM

 2x-1=14-x

2x+x=14+1

3x=15

x=15/3 

x=5

-Answered by Jyoti Srivastva On 07 July 2019:10:05:52 PM

 yes zero is a rational number.

-Answered by Himanshi Verma On 14 May 2019:11:23:16 AM

 4z+3=6+2z

4z-2z=6-3

2z=3

z=3/2

-Answered by Jyoti Srivastva On 07 July 2019:10:11:51 PM

 The rational number are 16, 17.

-Answered by Himanshi Verma On 14 May 2019:11:16:10 AM

 Density is important because it affects whether objects will float or sink. It is animportant property to consider when building things like ships and hot air balloons. ... He realized that the amount of water that spilled was equal in volume to the space that his body occupied.

-Answered by rocki kumar On 08 May 2019:08:23:43 AM

 Arrangement of metals in order of decreasing reactivities in vertical column is called activity series.

-Answered by Himanshi Verma On 21 April 2019:10:58:07 AM

When magnesium ribbon is burnt in oxygen it burns with a dazzling flame and gives a white magnesium oxide.

-Answered by Himanshi Verma On 21 April 2019:10:59:16 AM

 The reaction in which addition of oxygen and removal of hydrogen is called oxidation.

-Answered by Himanshi Verma On 21 April 2019:11:03:26 AM

9×LCM = 306×657

9×LCM = 201042

LCM   = 22338

-Answered by Himanshi Verma On 21 April 2019:10:32:51 AM

Let us assume that √5 is rational 

Then √5 =  

(a and b are co primes, with only 1 common factor and b≠0) 

⇒ √5 =  

(cross multiply) 

⇒ a = √5b 

⇒ a² = 5b² -------> α

⇒ 5/a² 

(by theorem if p divides q then p can also divide q²) 

⇒ 5/a ----> 1 

⇒ a = 5c 

(squaring on both sides) 

⇒ a² = 25c² ----> β 

From equations α and β 

⇒ 5b² = 25c²

⇒ b² = 5c² 

⇒ 5/b² 

(again by theorem) 

⇒ 5/b---> 2 

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong. 

This contradiction arises because we assumed that √5 is a rational number 

∴ our assumption is wrong 

∴ √5 is irrational number.

-Answered by Himanshi Verma On 21 April 2019:11:24:50 AM

 prime factors of 6ⁿ 

                = (2*3)ⁿ 

                = 2ⁿ*3ⁿ 

While the number ending with zero have prime factor as 2ⁿ*5ⁿ 

       Hence, 6ⁿ is not end with zero.

-Answered by Akki chauhan On 28 June 2019:09:30:20 AM

  Use Euclid division alogrithan 

 a=bq+r

225=135×1+90

135=90×1+45

90=45×2+0

HCF=45 .

-Answered by Himanshi Verma On 18 April 2019:04:26:31 PM

 Suppose, at first the side of the square was a

the area was a^2

now, the side of the square is (a+a★50/100)=a+a/2=3a/2

the area will be, 9a^2/4

so, 9a^2/4-a^2=5a^2/4

we now tell that, after increasing 50%of the side, the area will 5/4 times of the older(a^2).

-Answered by Himanshi Verma On 06 April 2019:06:44:55 PM

⇒% Decrease$\Rightarrow \mathrm{\%}Decrease$

=8−58×100=3712%v

-Answered by Himanshi Verma On 06 April 2019:06:46:28 PM

  LET THE ORIGINAL LENGTH BE L

AND

ORIGINAL BREATH BE B.

ORIGINAL AREA = LB

LENGTH INCREASED BY 50% = 150 L / 100

BREATH INCREASED BY 20% = 120 B / 100

INCREASED AREA = 150 L X 120 B / 100 X 100

= 18000 LB / 10000

= 1.8 LB

NOW,

1.8 LB / LB = 1.8

CHANGING 1.8 INTO FRACTION.

1.8 = 18/10 = 9/5 

SO,

THE INCREASED AREA WILL BE 9/5 TIMES OF THE ORIGINAL AREA.

-Answered by Hitesh kumar On 13 April 2019:08:31:51 AM

 so let us first consider a square with area x² unit²

so the side is of length √x² = x unit 

so square has angles of 90° so using Pythagoras we  get the length of the diagonal as  = 

√2x

so for this problem the area is 16900  m² so length of the side is √16900 = 130 m

so the length of the diagonal is √2 x 130 m = 130√2 m ≈ 183.82 m ANSWER

-Answered by Himanshi Verma On 06 April 2019:06:51:54 PM

 The perimeter is given by twice the sum of length and breadth.

-Answered by Himanshi Verma On 06 April 2019:06:53:50 PM

In the figure, ABCD is a rectangular field.
AC is the diagonal.
Triangle ABC, 
AC2=AB2+BC2$A{C}^2=A{B}^2+B{C}^2$
202=162+BC2${20}^2={16}^2+B{C}^2$
BC2=202−162$B{C}^2={20}^2-{16}^2$
=(20+16)(20−16)=36×4$\left(20+16\right)\left(20-16\right)=36\times 4$
BC=6×2$BC=6\times 2$ = 12 m.
The breadth of the field is 12 m.

-Answered by Himanshi Verma On 06 April 2019:06:43:16 PM

 (i)135 and 225

  a = 225, b = 135               {Greatest number is ‘a’ and smallest number is ‘b’}

 Using Euclid’s division algorithm

  a = bq + r (then)

  225 = 135 ×1 + 90

 135 = 90 ×1 + 45

 90 = 45 × 2 + 0                  {when we get r=0, our computing get stopped}

 b = 45 {b is HCF}

Hence:  HCF = 45

 (ii)196 and 38220

 a = 38220, b = 196          {Greatest number is ‘a’ and smallest number is ‘b’}

 Using Euclid’s division algorithm

 a = bq + r (then)

 38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}

 b = 196 {b is HCF}

Hence:  HCF = 196

(iii)867 and 255

a = 867, b = 255               {Greatest number is ‘a’ and smallest number is ‘b’}

Using Euclid’s division algorithm

a = bq + r (then)

38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}

b = 196 {b is HCF}

Hence:  HCF = 196

-Answered by Himanshi Verma On 29 March 2019:04:55:16 PM

 Vitamin D.

-Answered by Himanshi Verma On 23 March 2019:05:59:38 PM

 Since 2 is one root of the quadratic equation,

We substitute X=2

k(2)^2 - 14(2) +8= 0

4k-28+8=0

4k-20=0

4k=20

k=5

-Answered by Himanshi Verma On 10 March 2019:04:48:03 PM

 ABC is a right angle triangle at B and BD is a perpendicular from B on to AC, the diagonal.

Triangles ABC and BDC are similar, as:

     . angle DBC = 90 - angle C = angle A

     . angle ABC = angle BDC = 90

Ratios of corresponding sides are equal.

     AB / BD = BC / DC = AC / BC

    So we get  BC² = AC * DC     --- (1)

Triangle ABC amd ADB are similar, as:

    . angle DBA = 90 - (90 - C) = angle C

    . angle ABC = 90 = angle ADB

so ratios of corresponding sides are equal

         AB / AD = BC / BD = AC / AB 

       So we get  AB² = AD * AC  --- (2)

  Add (1) and (2) to get:

       AB² + BC² = AC * ( AD + DC) = AC * AC = AC²

So the theorem is proved.

-Answered by Himanshi Verma On 01 March 2019:03:38:26 PM

-Answered by Shalu Shukla On 01 March 2019:03:47:38 PM

 Let us draw a line segment AM ⊥ BC.

We know that,

Area of a triangle = 1/2 x Base x Altitude

It is given that DE = BD = EC.

⊥ Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)

It can be observed that Budhia has divided her field into 3 equal parts.

-Answered by Himanshi Verma On 09 February 2019:04:40:18 PM

 If an event is impossible its probability is zero. Similarly, if an event is certain to occur, its probability is one. The probability of any event lies in between these values. It is called the range of probability and is denoted as 0 ≤ P (E) ≤ 1.

-Answered by Himanshi Verma On 09 February 2019:04:32:27 PM

 बेलन के पानी का आयतन = गोले का आयतन 

   πr²h   = 4/3 πr³

   14×14×h = 4\3×7×7×7

          h   = 4/3 × 343/196

          h  = 7/3cm

-Answered by Shalu Shukla On 11 February 2019:05:45:38 PM

 Considering AB = Base;

Area = Base x Height

⇒ 1470 cm2 = AB x DL

⇒ 1470 cm2 = 35cm x DL

Or, DL = 1470 ÷ 35 = 42 cm

Now considering AD = Base;

Area = Base x Height

⇒ 1470 cm2 = AD x BM

⇒ 1470 cm2 = 49 cm x BM

Or, BM = 1470 ÷ 49 = 30 cm

Hence, DL = 42 cm and BM = 30 cm

-Answered by Himanshi Verma On 09 February 2019:04:21:59 PM

  8 8 8

     8 8

        8

        8

+      8

= 1000 

-Answered by Himanshi Verma On 05 February 2019:05:36:22 PM

 Given

Ratio of side =3:2

Perimeter =60cm

Altitude =5cm

So let take one side as 3x and other is 2x

Then perimeter of parallelogram 

=3x+2x+3x+2x

=10x 

According to question perimeter =60cm 

Then 

10x=60cm 

X=60/10=6cm

Then one side is =6 x 3=18

And other is =2x6=12

So base=18cm 

Area of parallelogram =bxh

=18x5=90cm²

Altitude corresponding to smaller side 

90=12xh

H=7.5cm 

-Answered by Himanshi Verma On 05 February 2019:05:26:39 PM

 Diameter is the greatest chord of circle.

-Answered by Himanshi Verma On 05 February 2019:04:58:06 PM

 9+8+8+7+7=39,

9+9+8+3=29,

7+2=9

We write last nine of the three solution so, we get 777+888+999=999 

Hence, proved.

-Answered by Nishant Verma On 03 February 2019:05:23:34 PM

 9+8+8+7+7=39,

9+9+8+3=29,

7+2=9

We write last nine of the three solution so, we get 777+888+999=999 

Hence, proved.

-Answered by Himanshi Verma On 05 February 2019:04:51:38 PM

 The usual definition of pi is the ratio of the circumference of a circle to its diameter, so that the circumference of a circle is pi times the diameter, or 2 pi times the radius. ... This give a geometric justification that the area of a circle really is "pi r squared".

-Answered by Himanshi Verma On 02 February 2019:01:33:01 PM

 (secA+tanA-(sec^2 A - tan ^2A)) )/tanA-secA+1) as sec^2 A= 1 + tan ^2 A so 1= sec^2 A - tan ^2 A 

= (sec A + tan A - (sec A + tan A) ( sec A - tan A)) / ( tanA-secA+1) 

= ( sec A + tan A ) ( 1- (sec A - tan A)/ ( tanA-secA+1) 

= (sec A + tan A)/(+ tan A - sec A)/(tan A- sec A+ 1) 

= (sec A + tan A) 

= 1/cos A + sin A/ cos A 

= (1+ sin A)/ cos A 

= (1 + sin A )(1- sin A)/(cos A (1- sin A)) 

= (1- sin ^2 A/(cos A (1- sin A)) 

= cos ^2 A / (cos A (1- sin A)) 

= cos A /(1- sin A) 

proved

-Answered by Himanshi Verma On 02 February 2019:01:40:14 PM

 Length (l) of garden = 90 m

Breadth (b) of garden = 75 m

Area of garden = l × b = 90 × 75 = 6750 m2

From the figure, it can be observed that the new length and breadth of the garden, when path is also included, are 100m and 85m respectively.

Area of the garden including the path = 100 × 85 = 8500 m2

Area of path = Area of the garden including the path − Area of garden

= 8500 − 6750 = 1750 m2

1 hectare = 10000 m2

Therefore, area of garden in hectare

=6750/10000

=0.675

-Answered by Himanshi Verma On 02 February 2019:01:49:33 PM

 A=6400m²

-Answered by Himanshi Verma On 03 February 2019:04:19:42 PM

 The given quadratic will be a perfect square if it has two real and equal roots. To have two real and equal roots the discriminant must be zero.

i.e., b²-4ac=0

or, (2k+4)²-4(4-k)(8k+1)=0

or, 4k²+16k+16-4(32k-8k²+4-k)=0

o², 4k²+16k+16-128k+32k²-16+4k=0

or, 36k²-108k=0

or, 36k(k-3)=0

either 36k=0

or, k=0

or, k-3=0

or, k=3

∴, k=0,3 Ans.

-Answered by Himanshi Verma On 02 February 2019:01:43:20 PM

 Consider, abx2 +(b2-ac)x-bc = 0

-Answered by Himanshi Verma On 02 February 2019:01:44:13 PM

ON solving the above equation we have a quadratic equation  

(a+b)x^2+(a+b)^2x+ab(a+b)

on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac)

we got -a and -b as roots.

According to me this is much easier then others when compared.

-Answered by Himanshi Verma On 10 March 2019:04:49:29 PM

 : (11 + 9) + (3) + (7) = 30

ANOTHER: (13 + 15) + (1) + (1) = 30

OR: (7 + 9) + (1) + (13) = 30

-Answered by Himanshi Verma On 21 January 2019:04:52:39 PM

 Cscϕ/secϕ = (1+cotϕ)/(1+tanϕ) 

LHS = (1+cotϕ)/(1+tanϕ) 

where cotϕ=cosϕ/sinϕ and tanϕ = sinϕ/cosϕ 

= (1+(cosϕ/sinϕ))/(1+(sinϕ/cosϕ)) 

= ((cosϕ+sinϕ)/sinϕ)/((cosϕ+sinϕ)/cosϕ) 

= cosϕ/sinϕ 

wkt cosϕ=1/secϕ 

and sinϕ=1/cosecϕ 

= cscϕ/secϕ 

LHS = RHS 

Hence Proved

-Answered by Master Purushottam On 24 August 2019:12:17:05 AM

 Total no. of times tossing a coin= 1000

no. of head comes= 425

no. of tail comes= 1000-425= 575

probability   p(e)= 575/1000 

-Answered by Master Purushottam On 24 August 2019:12:18:13 AM

Let three spheres are S1, S2 & S3

having radii r₁ = 3cm, r₂ = 4cm & r₃ = 5cm respectively.

Let the radius of new Big sphere S is R.

A/Q,

Volume of new Sphere  S = Sum of volumes three Spheres S1, S2 & S3

⇒ 4/3πR³ = 4/3π(r₁)³ + 4/3π(r₂)³+4/3π(r₃)³

⇒ 4/3πR³ = 4(r₁³ +r₂³ + r₃³)/3π

⇒ R³ = r₁³ +r₂³ + r₃³

⇒  R³ = 3³ +4³ + 5³ = 27 + 64 + 125

⇒ R³ =  216

⇒ R = ∛216

⇒ R = 6cm

Therefore radius of new Sphere is 6 cm.

-Answered by Master Purushottam On 24 August 2019:12:18:19 AM

 slant height of cone is : 10 

Radius of the base is : 8 cm

Total surface area of cone is : πr(l+r)

227 ×8(10+8) 

227 ×144

= 452.57

-Answered by Master Purushottam On 24 August 2019:12:15:34 AM

 Total surface area= 2πr(r+h) -πr²

curved surface area = 2πrh+πr²

-Answered by Master Purushottam On 24 August 2019:12:15:48 AM

 vb

-Answered by Master Purushottam On 24 August 2019:12:14:15 AM

  Let ABCD, the square garden of side= 30 m.

PQRS is the region inside the garden.

so,

PQ = (30 -1-1 ) = 28 m

also

PS = ( 30 – 1 – 1 ) = 28 m 

(i) Area of the path = Area of ABCD – Area of PQRS 

= [( 30 x 30) – (28 x 28)]

= 900 – 784 

= 116 sq.m 

(ii) Area of the remaining portion in which the grass is planted = Area of square PQRS 

= 28 x 28 

= 784 sq.m 

cost of planting the grass in the region PQRS =  area x cost = 764 x 40

 =31,360

-Answered by Master Purushottam On 24 August 2019:12:14:42 AM

-Answered by Master Purushottam On 24 August 2019:12:15:00 AM

-Answered by Master Purushottam On 24 August 2019:12:15:06 AM

 We know,

---> if a + b + c = 0 ; a³ + b³ + c³ = 3abc

We observe,

( a² - b² ) + ( b² - c² ) + ( c² - a² ) = 0

=> ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ = 3( a² - b² )( b² - c² )( c² - a² )

Similarly, 

( a - b ) + ( b - c ) + ( c - a ) = 0

=> ( a - b )³ + ( b - c )³ + ( c - a )³ = 3( a - b )( b - c )( c - a )

Now, [ ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ ] / [ ( a - b )³ + ( b - c )³ + ( c - a )³ ]

== 3( a² - b² )( b² - c² )( c² - a² ) / 3( a - b )( b - c )( c - a )

-Answered by Master Purushottam On 24 August 2019:12:09:53 AM

 2

-Answered by Master Purushottam On 24 August 2019:12:11:09 AM

-Answered by Master Purushottam On 24 August 2019:12:11:22 AM

  Lateral Surface Area of cube=256 cm²

-Answered by Master Purushottam On 24 August 2019:12:03:45 AM

 l=5m    b=4m      h=3m 

The total area of cuboid = 2(lb+bh+hl)+l×b

                                  = 2(5×4+4×3+3×5)+5×4

                                  = 2(20+12+15)+20

                                  = 2(47) +20

                                  =    94+20

                                  =     114m²

The cost of white washing the wall and the ceiling of the room = 7.50 per m²

      The cost of 114m² of the walls and ceiling of the room= (7.50×114)Rs

                                                                                   = 855.00Rs.

-Answered by Master Purushottam On 24 August 2019:12:04:01 AM

The total surface area of hemisphere= 3πr²

The ratio of total surface area of sphere and hemisphere= 4πr²/3πr²

                                                                                = 4:3

-Answered by Master Purushottam On 24 August 2019:12:04:08 AM

   Total surface area of cube is 150sq. cm

Total surface area of cube : 6a²

^{To find perimeter of its one faces :  ?}

Total surface area of cube =150sq.cm

                           6a²  =150sq.cm

 = 1506 

= 25cm

=a² = √25cm

=a = 5 cm

Perimeter of its one face is = 4 × a

                                        = 4 × 5 cm

                                        = 20 cm Answer

-Answered by Master Purushottam On 24 August 2019:12:01:30 AM

The surface area of the cuboid= 1372 sq. cm 

  Let the dimensions =4x, 2x, 1x

     The total surface area of the cuboid = 2(lb+bh+hl)=1372

                                                        = 2(4x × 2x + 2x × 1x + 4x × 1x )=1372 

=2(8x²+2x²+4x²) =1372

=2(14x²)= 1372

=28x²=1372

= x²=1372/28

       = 49

x=7m

the dimensions are 4x=4×7=28m

                           2x=2×7=14m

                           1x=1×7=7m

-Answered by Master Purushottam On 24 August 2019:12:01:38 AM

 Let the radii be 2x and 3x,

let the height be 5y and 3y.

so, ratio of volume =r²h/R²H =(2x)²×5y/(3x)²×3y =20x²y/27x²y =20/27,

The ratio is = 20:27

-Answered by Master Purushottam On 23 August 2019:11:57:01 PM

 Let the first and second angle be a and b respectively 

-Answered by Master Purushottam On 23 August 2019:11:41:59 PM

 We know, 

-Answered by Master Purushottam On 23 August 2019:11:36:15 PM

  Ans 1. Zero(0)

Ans 2. 0 and 1

Ans 3. 2

-Answered by Master Purushottam On 23 August 2019:11:36:37 PM

Let 'x' be the number of boys and 'y' be the number of girls.

To find the total marks

The total marks of boys = 42x

The total marks of girls = 45y

Total number of boys and girls = x + y

To find the ratio

The total marks of boys and girls = 44(x + y)

Then we can write

44(x + y) = 42x + 45y

44x + 44y = 42x + 45y

2x = y

x/y = 1/2

Therefore the ratio of the number of boys to the number of girls is 1:2

-Answered by Master Purushottam On 23 August 2019:11:37:23 PM

 Let 'x' be the number of boys and 'y' be the number of girls.

To find the total marks

The total marks of boys = 42x

The total marks of girls = 45y

Total number of boys and girls = x + y

To find the ratio

The total marks of boys and girls = 44(x + y)

Then we can write

44(x + y) = 42x + 45y

44x + 44y = 42x + 45y

2x = y

x/y = 1/2

Therefore the ratio of the number of boys to the number of girls is 1:2

-Answered by Master Purushottam On 23 August 2019:11:37:42 PM

   l=10cm r=8cm

The total surface area of cone=πr(l+r)

                                          =22/7×8(10+8)

                                          =22/7×8(18)

                                          =22/7×144

                                          =3168/7cm

-Answered by Master Purushottam On 23 August 2019:11:37:53 PM

r= 32

Volume of hemisphere= 23 πr³

                                = 23 ×22/7×3/2×3/2×3/2

                                 = 7.071

-Answered by Master Purushottam On 23 August 2019:11:33:27 PM

-Answered by Master Purushottam On 23 August 2019:11:34:12 PM

 √125 +2√27 +(-5√5 -√3)

=√5×5×5 +2√3×3×3 -5√5 -√3 

=5√5 +6√3 -5√5 -√3

=5√3

-Answered by Akki chauhan On 15 January 2019:09:05:29 AM

  ∠A + ∠C = 180 ....1

But ∠A = ∠C

So ∠A = ∠C = 90

Again 

  ∠B + ∠D = 180 ....2

But ∠B = ∠D

So ∠B = ∠D = 90

Now each angle of parallelogram ABCD is 90.

Hense ABCD is a rectangle

-Answered by Akki chauhan On 15 January 2019:09:05:05 AM

2x-5=0

2x=5

x= 5/2

-Answered by Master Purushottam On 22 August 2019:09:11:57 PM

 x³ -23x²+142x-120 

=x³-x²-22x²+22x+120-120

=x²(x-1)-22x(x-1)+120(x-1)

=(x-1)(x²-22x+120)

=(x-1)(x²-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

-Answered by Master Purushottam On 22 August 2019:09:12:12 PM

When p(x) is divided by (x-2),

By remainder theorem,

Remainder= p(2)

p(2)=a*2^3 + 3*2^2 - 13

        =a*8+3*4-13

       =8a+12-13 = 8a-1

p(2) = 2*2^3-5*2+a

      =2*8-10+a

     =16-10+a = 6+a

Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,

8a-1 = 6+a

8a-6+a=1

8a+8=1-6=-5

8a=-5-8 =-13

a=-13/8

-Answered by Master Purushottam On 22 August 2019:09:12:19 PM

 (a+1/a)2=a2+1/a2+2

a2+1/a2=(a+1/a)2-2→ I

Now,

a=9-4root5

1/a=1/9-4root5

=(1/9-4root5)×9+4root5/9+4root5

=(9+4root5)

a+1/a=9-4root5+9+4root5

=81

Now,

a2+1/a2=(a+1/a)2-2 →(from I)

=(81)2-2

=6561-2

=6559

-Answered by Master Purushottam On 22 August 2019:09:08:56 PM

 The first 10 prime numbers are : 2,3,5,7,11,13,17,19,23 and 29.

n=10

Median if n is even numbers

n2 th term = 5th term = 11

n2 + 1 = 5 + 1 = 6th term = 13

Median = 5th term + 6th term2 

Median = 11 + 132 

 = 242 = 12 Answer

-Answered by Master Purushottam On 22 August 2019:09:04:22 PM

-Answered by Master Purushottam On 22 August 2019:09:04:32 PM

volume of the sphere = 43 πr³= 310.4cm³

                                4/3×22/7×r³=310.4

                                88/21×r³= 310.4

                                   r³=   310.4×21/88

                                      =   6518.4/88

                                   r³  = 74.0727272cm

-Answered by Master Purushottam On 22 August 2019:09:04:44 PM

 Highest power of x in the algebraic expression is called the degree of the polynomials.

-Answered by Prince Rawat On 11 January 2019:10:08:00 AM

Given : In a circle C(O,r) , ∠AOB = ∠COD 

To Prove : Chord AB = Chord CD .

Proof : In △AOB and △COD 

AO = CO [radii of same circle]

BO = DO [radii of same circle]

∠AOB = ∠COD  [given]

⇒  △AOB ≅  △COD [by SAS congruence axiom]

⇒   Chord AB = Chord CD [c.p.c.t]

-Answered by Master Purushottam On 22 August 2019:08:59:15 PM

  Start with: -20 = -20

-Answered by Master Purushottam On 22 August 2019:09:00:11 PM

The basis of the Euclidean division algorithm isEuclid's division lemma. To calculate the Highest Common Factor (HCF) of two positive integers a and b we use Euclid's division algorithm. HCF is the largest number which exactly divides two or more positive integers.

-Answered by Master Purushottam On 22 August 2019:09:00:20 PM

  21b - 32 + 7b - 20b

=(21b+7b -20b) - 32

=8b - 32 Answer

-Answered by Master Purushottam On 22 August 2019:09:00:32 PM

 3p²q² - 4pq + 5 + (-10p²q²) + 15 + 9pq + 7p²q² 

= 3p²q² + 7p²q² - 10p²q² + 9pq - 4pq + 5 + 15 

= 10p²q² - 10p²q² + 5pq + 20 

= 5pq + 20

-Answered by Master Purushottam On 22 August 2019:09:00:43 PM

 ( x + 1/x)² = x² + 1/x² + 2

=> 2² = x² + 1/x² + 2

=> x² + 1/x² = 4–2 = 2 . . . . . . . .(1)

(x - 1/x)² = x² + 1/x² -2

=> ( x-1/x)² = 2 - 2 = 0

=> (x-1/x) = 0 . . . . . . . . . .(2)

Now, x^4 - 1/x^4 = ( x² + 1/x²) ( x² - 1/x²)

= 2 ( x + 1/x) ( x-1/x)

= 2 * 2 * 0 ( by (1) & (2)

= > x^4 - 1/x^4 = 0 

-Answered by Himanshi Verma On 22 March 2019:05:16:46 PM

                   =4

The value of b²= (-2²) = 4

 a²+b²= 4+4

         =  8

-Answered by Master Purushottam On 22 August 2019:08:55:06 PM

 x+7+4(x-5)=2

x+7+4x-20=2

5x-13=2

5x=2+13

5x=15 

x=3

So, The value of x is 3. 

-Answered by Master Purushottam On 22 August 2019:08:55:13 PM

 a=5,      b=-3

2(5²+5×(-3))+3-5×(-3)

2(25+(-15))+3+15

2(25-15)+18

2×10+18

20+18

38

-Answered by Master Purushottam On 22 August 2019:08:55:23 PM

 3a-2b-ab-(a-b+ab)+3ab+b-a

-Answered by Master Purushottam On 22 August 2019:08:55:31 PM

-Answered by Master Purushottam On 22 August 2019:08:55:50 PM

    x² + 2x + 1 

= x² + x + x + 1

= x(x + 1) + 1(x + 1) 

= (x + 1) (x + 1) 

-Answered by Akki chauhan On 10 January 2019:10:57:10 AM

 For finding the value of x first we find mean of 5, 10, 3 

sum of observation = 5 + 10 + 3 = 18

mean = 18 / 3 = 6 

Therefore, x = 6 

Now we have data for median 11, 12, 14, 18, 21, 23 

n = 6 { n is an even number} 

n/2 th trem = 14 

and (n/2 + 1)th trem = 18 

Median = (14 + 18)/2 = 32/ 2 = 16 Answer

-Answered by Master Purushottam On 22 August 2019:08:53:15 PM

S7 = 7/2(2a + 6d) 

=> 49 = 7a + 21d

=> 7 = a + 3d   ------------ (i) 

S17 = 17/2 (2a + 16d) 

=> 289 = 17a + 136

=> 17 = a + 8d --------------  (ii) 

Substracting equation (i) from (ii) 

we have ... 

17 - 7 = a - a + 8d - 3d 

10 = 5d 

d = 10/5 

d = 2 

Putting the value in Equation (i) 

7 = a + 3d 

7 = a + 3 x 2 

a = 7 - 6 

a = 1 

Sum of first n terms = n/2 [2a + (n - 1) d]

= n/2 [2x 1 + (n - 1)2] 

= n/2 [2 + 2n - 2 ]

= n/2 (2n) 

= nxn = n2

-Answered by Master Purushottam On 22 August 2019:08:54:07 PM

 A.p= 9,17,25

a=9

d=17-9

=8

Sum of n terms=n÷2[2a+(n-1)d

636=n÷2[2(9)+(n-1)8]

636=n÷2[18+8n-8]

636=n÷2[10+8n]

1272=10n+8n^2

8n^2+10n-1272=0

2[4n^2+5n-636]=0

4n^2+5n-636=0

4n^2+53n-48n-636=0

n(4n+53)-12(4n+53)=0

(n-12)(4n+53)=0

n-12=0

n=12

12 terms must be given to sum of 636

-Answered by Master Purushottam On 22 August 2019:08:54:23 PM