CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Let rate of bat = xRs. and rate of ball = yRs. 3x+6y=3900.....(i) and x+3y=1300.....(ii) or x=(1300−3y) now equation (i) can be written as 3(1300−3y)+6y=3900 3900−9y+6y=3900 ∴y=0 Ans:- x=1300

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Real no. Theorem 1.4 in brief

Posted by Shardul Rai 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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The HCF (a, b) = 2 and LCM (a, b) = 27. What is the value a × b

Posted by Anshu Dudi 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

The value of a*b is 54...hope it would be helpful✌

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Anshu Dudi 5 years, 1 month ago

PLEASE ANSWER IT FAST AS TOMORROW IS MY PAPER

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In a triangle ABC, write cos (B+C/2) in terms of Angel a

Posted by Deepak Karthick 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Utsav Kumar 5 years, 1 month ago

We know that, A+ B+ C = 180° B+C = 180° - A Dividing both sides by 2 B+C /2 = 90° - A/2 Taking cos on both sides, cos(B+C/2) = cos(90° - A/2) This gives, cos(B+C/2)= sin(A/2)

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In fig. PQ and PR are tangents to circle with centre O such that angleQPR =50

Posted by Riddhi Bishnoi 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 5 years ago

In quad. PQOR,
∠R+ ∠O+ ∠Q+ ∠P = 360
∠P + ∠O = 90
∠O = 130
Now,
In ∆ROQ,
∠RQO + ∠QOR + ∠ORQ = 180°
Now as the sides OQ = RO because of radius of same circle... Hence.... OQR = 25°

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A dice is tossed find the probability of getting a number is even or a multiple of 3

Posted by Ritika Ritika 5 years, 1 month ago

CBSE > Class 10 > Mathematics

5 answers

Meenakshi Somanchi 5 years, 1 month ago

1/6 Reason:- A number which is both an even and a multiple of 3 is 6 Therefore, P(E) = number of possible outcomes/ total number of outcomes =》 P(E) 1/6

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Anuj Kumar 5 years, 1 month ago

2 by 6 kaise aayega

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Prachi Jain 5 years, 1 month ago

2 by 3

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Anuj Kumar 5 years, 1 month ago

4 by6 = 2 by 3

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Account Deleted 5 years, 1 month ago

2 by 6

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Which of the following are AP ? If they form an AP, find the Common Difference D and write more 3 terms. (i) a, a², a³, a⁴.... (ii)1², 3², 5², 7²..... (iii) √2, √8, √18, √32.....

Posted by Manish Agarwal 5 years, 1 month ago

CBSE > Class 10 > Mathematics

5 answers

........ ...... 5 years, 1 month ago

You're mos welcome ?

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Manish Agarwal 5 years, 1 month ago

Thank you Saanvi

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........ ...... 5 years, 1 month ago

Option i) and ii) are not forming AP. iii)D=second term - first term Here the common difference is, a2-a1= 2√2-√2=√2 a3-a2=3√2-2√2=√2 a4-a3=4√2-3√2=√2 Since, the D is same every time . It is an AP. :. D= √2 And 3 more terms are a5= 4√2+√2=5√2=√50, a6=5√2+√2=6√2=√72, a7=6√2+√2=7√2=√98

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Naveen Dhongade 5 years, 1 month ago

Hvvb

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Ankit Kumar 5 years, 1 month ago

Hui

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If a(3,y) is equidistant from p (8,-3) q (7, 6) find the value of y

Posted by Yashvi Siwan 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

आदित्य बड़गुर्जर 5 years, 1 month ago

pa=qa pa= √ (8-3)²+(-3-y)²= qa =√(7-3)²+(6-y)² √25+y² +6y+9=√16+y²+12y+36 34+y²+6y = 52 +y² +12y 6y = -18 y = -18/6 = -3 Hence y=-3

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Find the greatest number which leaves a reminder 10 in each case when divided by 56,84 and 140

Posted by Đivyansh Chaurasia 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Zeroes of quadratic polynomial and verify relationship between zeroes and coefficients 4u²+8u

Posted by Nancy Lakhlan 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Prachi Jain 5 years, 1 month ago

Here, 4u^2+8u=0 =4u(u+2)=0 4u=0 or u+2=0 U=0 or u=-2 :. Zeroes of the given quadratic equation is 0 and -2 Then, alpha=0 and beta=-2 Where, a=4, b=8, c=0 The relationship between zeroes and coefficient of given quadratic equation: Alpha+beta=-b/a :.0+(-2)=-8/4 -2=-2 Alpha×beta=c/a :.0×-2=0/4 0=0

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........ ...... 5 years, 1 month ago

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For the following APs, write the first term and the common difference: (i) 3, 1, – 1, – 3,...

Posted by Sahil Routhan 5 years, 1 month ago

CBSE > Class 10 > Mathematics

3 answers

Rudraksh Rai Udaiwal 5 years, 1 month ago

a=3, d=-2

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Manish Agarwal 5 years, 1 month ago

First term (a) :- 3 D :- a²-a¹ = 1-3 = -2

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Mayank Chaudhary 5 years, 1 month ago

a=3 ; d= a2- a1=1-3=-2

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Use Euclid division algorithm to find the HCF of 135and225

Posted by Affan Ahmad 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

........ ...... 5 years, 1 month ago

By Euclid division algorithm, a=bq+r, 225>135 225=135×1+90 135=90×1+45 90=45×2+0 Since, r=0 :. HCF of 135 and 225 is 45

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What is digree of cubic polynomial

Posted by Aarush Arora 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Simranjeet Yadav 5 years, 1 month ago

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Tamanna Rathee 5 years, 1 month ago

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Neha Roy 5 years, 1 month ago

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Roopam Almadi 5 years, 1 month ago

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Aarush Arora 5 years, 1 month ago

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For what value k, will the system of equations X+2y=5 3x+ky+15=0 Have a unique solution

Posted by Aravind S 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Roopam Almadi 5 years, 1 month ago

K is not equal to 6 For all the values of k other than 6 The equation has a unique solution

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........ ...... 5 years, 1 month ago

Here, a1=1,b1=2,c1=-5 a2=3, b2=k, c2=15 For a pair of linear equation to have a unique solution:a1/a2 not equal to b1/b2 =1/3 not equal to 2/k ( solve this equation) ,we get K=6 :.The value of k=6

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1. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr. 2. Two ships are sailing in the sea on the either side of the light-house, the angles of depression of two ships as observed from the top of the light-house are 60° and 45° respectively. If the distance between the ships is 200 metres, find the height of the light-house. 3. An aeroplane, when 1500 m high passes vertically above another aeroplane at an instance when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two aeroplanes.

Posted by Anchør Skull Chalengers 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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Chapter 3,4,5 ncert solutions in very easy way

Posted by Yashika Sharma 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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The circle is touching the side BC of ABC at p and touching AB and AC produced at Q and R respectively proved that AQ 1/2 (perimeter of ABC)

Posted by Faizan Hussain 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

☆•..¤( Prateek )¤..•☆ 5 years, 1 month ago

Given: A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R respectively. RTP: AP = 1/2 (Perimeter of ΔABC) Proof: Lengths of tangents drawn from an external point to a circle are equal. ⇒ AQ = AR, BQ = BP, CP = CR. Perimeter of ΔABC = AB + BC + CA = AB + (BP + PC) + (AR – CR) = (AB + BQ) + (PC) + (AQ – PC) [AQ = AR, BQ = BP, CP = CR] = AQ + AQ = 2AQ ⇒ AQ = 1/2 (Perimeter of ΔABC) ∴ AQ is the half of the perimeter of ΔABC Hope it will help you ?✌

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Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

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Volume of cylinder

Posted by Farid Khan 5 years, 1 month ago

CBSE > Class 10 > Mathematics

3 answers

Yogita Ingle 5 years, 1 month ago

A cylinder can be seen as a collection of multiple congruent disks, stacked one above the other. In order to calculate the space occupied by a cylinder, we calculate the space occupied by each disk and then add them up. Thus, the volume of the cylinder can be given by the product of the area of base and height.

For any cylinder with base radius ‘r’, and height ‘h’, the volume will be base times the height.

Therefore, the cylinder’s volume of base radius ‘r’, and height ‘h’ = (area of base) × height of the cylinder

Since the base is the circle, it can be written as

Volume = πr²× h

Therefore, the volume of a cylinder = πr²h cubic units.

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Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

I mean πr^2h

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Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

πr^2

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What is f of x

Posted by Siddharth Kumar ✔️✔️✔️ 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

☆•..¤( Prateek )¤..•☆ 5 years, 1 month ago

Yes, please write complete question. ?✌

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Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

Please write complete question. ?✌

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For any positive integer n,prove that n^3-n is divisible by 6

Posted by Aadya Singh 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 5 years ago

Answer:

there are two methods to solve the problem which are discussed below.

Method 1:

Let us consider

a = n³ – n

a = n (n² – 1)

a = n (n + 1)(n – 1)

Assumtions:

1. Out of three (n – 1), n, (n + 1) one must be even, so a is divisible by 2.

2. (n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3.

From (1) and (2) a must be divisible by 2 × 3 = 6

Thus, n³ – n is divisible by 6 for any positive integer n.

Method 2:

When a number is divided by 3, the possible remainders are 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where r is some integer.

Case 1: Consider n = 3p

Then n is divisible by 3.

Case 2: Consider n = 3p + 1

Then n – 1 = 3p + 1 –1

⇒ n -1 = 3p is divisible by 3.

Case 3: Consider n = 3p + 2

Then n + 1 = 3p + 2 + 1

⇒ n+1 = 3p + 3

⇒ n+1 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.

⇒ n (n – 1) (n + 1) is divisible by 3.

Similarly, when a number is divided by 2, the possible remainders are 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

Case 1: Consider n = 2q

Then n is divisible by 2.

Case 2: Consider n = 2q + 1

Then n–1 = 2q + 1 – 1

n – 1 = 2q is divisible by 2 and

n + 1 = 2q + 1 + 1

n +1 = 2q + 2

n+1= 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.

∴ n (n – 1) (n + 1) is divisible by 2.

Since, n (n – 1) (n + 1) is divisible by 2 and 3.

Therefore, as per the divisibility rule of 6, the given number is divisible by six.

n³ – n = n (n – 1) (n + 1) is divisible by 6.

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solve graphically and find the coordinates of the vertixes of the triangle.a)y=x y=2x x+y=6

Posted by Prasidh A Praveen 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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2/√x

Posted by Khushal Gurubakhshani 5 years, 1 month ago

CBSE > Class 10 > Mathematics

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from a point Q, the length of the tangets to a circle with centre O so that angle POQ=110°,then angle PTO is equal to

Posted by Teba Habib 5 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 5 years ago

It is given that,
∠POQ =110°
Since, the Iangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = 90°
and ∠OQT = 90°
Now, in quadrilateral POQT.
∠POQ + ∠OQT + ∠PTQ + ∠OPT = 360°
(angle sum property of quadrilateral)
⇒110° + 90° + ∠PTQ + 90° = 360°
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 70°

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By gometrical method find the all trigometrical ratios of 0°and90° using the properties of gemetry and trigometry

Posted by Vishal Kumar 5 years, 1 month ago

CBSE > Class 10 > Mathematics