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  • 2 answers

Yogita Ingle 6 years, 10 months ago

245 = 5 × 7 × 7
285 = 3 × 5 × 19
Therefore, HCF = 5

Kratika Nyati 6 years, 10 months ago

HCF(245,285)=5
  • 0 answers
  • 1 answers

Sia ? 6 years, 6 months ago

Let n = 4q + 1 (an odd integer)
{tex}\therefore \quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}
{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \quad \text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}
{tex}= 16{q^2} + 8q{/tex}
{tex}= 8 \left( 2 q ^ { 2 } + q \right){/tex}
= 8m, which is divisible by 8.

  • 1 answers

Sonam Sharma 6 years, 10 months ago

Let the piece of cloth is x metre and it costs Rs.y/metre. Then by the given conditions - ∴, xy=200 -------------(1) and (x+5)(y-2)=200 or, xy+5y-2x-10=200 or, 200+5y-2x-10=200 or, 5y-2x-10=0 or, 5y=2x+10 or, y=(2x+10)/5 Putting in (1), x(2x+10)/5=200 or, x(2x+10)=200×5 or, 2x²+10x=1000 or, x²+5x-500=0 or, x²+25x-20x-500=0 or, x(x+25)-20(x+25)=0 or, (x+25)(x-20)=0 either, x+25=0 or, x=-25 or, x-20=0 or, x=20 Since the length of the cloth can not be negative, ∴, x=20 metre Putting in (1), 20y=200 or, y=10 ∴, the cloth is 20 m long and the original rate is Rs. 10/metre.
  • 2 answers

Khushboo Giri 6 years, 10 months ago

Let √5be a rational number say p/q where q is not equal to zero and p and q are coprime. √5=p/q Squaring both side 5=p²/q² 5q²=p²----------(1) 5q²is divisible by 5 p² is divisible by 5 p is divisible by 5 let p=5a putting the value in equation 1 5q²=p² 5q²=(5a)² 5q²=25a² q²=5a² 5a² is divisible by 5 q² is divisible by 5 q is divisible by 5 Therefore both p&q have common factor 5 We supposed p&q are coprime. Our supposition is wrong √5 is irrational.

Yogita Ingle 6 years, 10 months ago

let root 5 be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
root 5=p/q
=> root 5 ×q = p
squaring on both sides
=> 5×q×q = p×p  ------> 1
p×p is divisible by 5
p is divisible by 5
p = 5c  [c is a positive integer] [squaring on both sides ]
p×p = 25c×c  --------- > 2
sub p×p in 1
5×q×q = 25×c×c
q×q = 5×c×c
=> q is divisble by 5
thus q and p have a common factor 5
there is a contradiction
as our assumsion p &q are co prime but it has a common factor
so √5 is an irrational

  • 2 answers

Ram Kushwah 6 years, 10 months ago

AB=AC

AB^2=AC^2

(0-3)^2+(2-p)^2=(o-p)^2+(2-5)^2

9+4+p^2-4p=p^2+9

-4p=-4

p=1

Honey ??? 6 years, 10 months ago

P=1
  • 2 answers

Ram Kushwah 6 years, 10 months ago

Pl check your question is not complete.

If roots are equal then

b^2=4ac

2^2=4 x p x -3

p=-1/3

 

 

Dia Khurana@1608 6 years, 10 months ago

Px x x + 2x - 3 Px sq. + 2x -3 You'll get a quadratic eq. Now u can use quadratic formula, middle term split and completing the square Out of three u can use any method for finding the value of x I hope u got ur answer....?
  • 2 answers

Helping Hand✋? 6 years, 10 months ago

16......❌ ........67.....✔️

Affu 😊 6 years, 10 months ago

Wrong .... Its answer is 67
  • 3 answers

Aditya Singh 6 years, 10 months ago

Sm:Sn=m^2:n^2 Sm/Sn=m/2(2a+(n-1)d)/n/2(2a+(n-1)d ) Sm/Sn=(2a+(m-1)d)/(2a+(n-1)d) =m^2/n^2×n/m=m/n 2a+(m-1)d/2a+(n-1)d=m/n By cross multiplication 2an+n(m-1)d=2am+m(n-1)d 2an+nmd-nd=2am+mnd-md 2an-nd=2am-md 2an-2am=nd-md 2a(n-m)=d(n-m) 2a=d am/an=a+(m-1)d/a+(n-1)d am/an=a+(m-1)2a/a+(n-1)2a am/an=a+2am-2a/a+2an-2a am/an=2am-a/2an-a am/an=a(2m-1)/a(2n-1)=2m-1/2n-1

Gaurav Seth 6 years, 10 months ago

Let Sm and Sn be the sum of the first m and first n terms of the A.P. respectively. Let, a be the first terms and d be a common difference.



⇒[2a+(m−1)d]
n =[2a(n−1)d]m
⇒2an+mnd−nd=2am+mnd−md
⇒md−nd=2am−2an
⇒(m−n)d=2a(m−n)
⇒d=2a
Now, the ratio of its mth and nth terms is


Thus, the ratio of its mth and nth terms is 2m – 1 : 2n – 1.

Divanshu Rawat 6 years, 10 months ago

Freinds please answer it
  • 1 answers

Khushboo Giri 6 years, 10 months ago

an=4 a+(n-1)d=4 a+2n-2=4 a+2n=6 a=6-2n-------(1) Sn=-14 n/2(2a+(n-1)d)=-14 n/2(2a+(n-1)2)=-14 n(a+n-1)=-14 n(6-2n+n-1)=-14-----from (1) n(5-n)=-14 5n-n²=-14 n²-5n-14=0 n²-(7-2)n-14=0 n²-7n+2n-14=0 n(n-7)+2(n-7)=0 (n+2)(n-7)=0 Therefore, n=7 Putting the value of n in equation (1) a=6-2×7 a=6-14 a=-8
  • 2 answers

Vikram Singh 6 years, 10 months ago

Cross multiply Ull get 3sinA +3cosA=5SinA -5cosA Transpose 3cosa + 5cosa=5sina -3sina 8cosa=2sina 8/2=sina/cosa 4=tan A The answer woul be 28 +2/8+7 30/9

Navjot Kaur 6 years, 10 months ago

plz friends answer this question
  • 1 answers

Saurabh Patel 6 years, 10 months ago

22
  • 3 answers

Yogita Ingle 6 years, 10 months ago

Median

  1. A line segment passing through a vertex to the midpoint of an opposite side is called a median.
  2. The point where the 3 medians meet is called the centroid of the triangle.
  3. Each median of a triangle divides the triangle into two smaller triangles which have equal area.


Altitude

  1. An altitude passing through a vertex and is at right-angle at the opposite side.
  2. The point where the 3 altitudes meet is called the ortho-center of the triangle.
  3. The altitude of a triangle may lie inside or outside the triangle.

Babji Dandu 6 years, 10 months ago

Median divides the triangle into two equal parts whereas altitude makes 90 degrees

Laxmikant Dhanka 6 years, 10 months ago

Altitude is always perpendicular or inclined at right angle but median is a line which divide aline in ratio 1:1
  • 3 answers

Kratika Nyati 6 years, 10 months ago

You need to have a solution then you may find value of k

Diksha Rana 6 years, 10 months ago

Please give answer with complete solution

Ajay Dhanush 6 years, 10 months ago

1
  • 2 answers

Khushboo Giri 6 years, 10 months ago

Let a be the odd positive intezer and b=8. By Euclid division lemma where r=0,1,2,3,4,5,6,7. When r=0 a=8q+0 When r=1 a=8q+1 (odd) When r=2 a=8q+2 When r=3 a=8q+3 (odd) When r=4 a=8q+4 When r=5 a=8q+5 (odd) When r=6 a=8q+6 When r=7 a=8q+7 (odd)

Yogita Ingle 6 years, 10 months ago

According to Euclid division lemma , a = bq + r where 0 ≤ r < b
Here we assume b = 8 and r ∈ [1, 7 ] means r = 1, 2, 3, .....7
Then, a = 8q + r
Case 1 :- when r = 1 , a = 8q + 1
squaring both sides,
a² = (8q + 1)² = 64²q² + 16q + 1 = 8(8q² + 2q) + 1
= 8m + 1 , where m = 8q² + 2q
case 2 :- when r = 2 , a = 8q + 2
squaring both sides,
a² = (8q + 2)² = 64q² + 32q + 4 ≠ 8m +1 [ means when r is an even number it is not in the form of 8m + 1 ]
Case 3 :- when r = 3 , a = 8q + 3
squaring both sides,
a² = (8q + 3)² = 64q² + 48q + 9 = 8(8q² + 6q + 1) + 1
= 8m + 1 , where m = 8q² + 6q + 1
You can see that at every odd values of r square of a is in the form of 8m +1
But at every even Values of r square of a isn't in the form of 8m +1 .
Also we know, a = 8q +1 , 8q +3 , 8q + 5 , 9q +7 are not divisible by 2 means these all numbers are odd numbers
Hence , it is clear that square of an odd positive is in form of 8m +1.

 

  • 1 answers

Raj Dhali 6 years, 10 months ago

Required volume of cuboidal pond =50×44×21/100 =462cubic metre Speed of flowing water=15km/hr =15000m/hr Let the time taken be x hrs. Distance covered by flowing water=15000x Diameter =14cm Radius=7/100m =0.07m Since, volume of ? flowed in pipe=volume of cuboidal pond πr^2h=462 22/7 *(0.07)^2 *15000x=462 231x=462 X=2hr
  • 2 answers

Kitzz Perey 6 years, 10 months ago

And thankx for such question

Kitzz Perey 6 years, 10 months ago

To prove OP os perpendicular bisector of AB in triangle PAC and PBC PA=PB (length of tangent drawn from an xterior point is equal) <PAC=<PBC (through equal sides ) CP=CP through SAS congruence AC=BC cpct Triangle ACP=BCP <ACP=<BCP <ACP+<BCP=18O°(linear pair) 2<ACP=180° <ACP=90°=<BCP
  • 4 answers

Rajput Singh 6 years, 10 months ago

₹21000

Anjali Mahlawat 6 years, 10 months ago

I understood. Let his salary be x, So, 18/100.x=3780 X=3780.100/18=21000rupee

Anjali Mahlawat 6 years, 10 months ago

How?

Kuldeep Prajapati 6 years, 10 months ago

₹21000
  • 5 answers

Jais Ravi 6 years, 10 months ago

For right angle triangle :- 1/2base×height For equilateral triangle :- √3/4 side^2 For any triangle :- √s(s-a)(s-b)(s-c) And |1/2 x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

Ritika Bisht 6 years, 10 months ago

1/2[x1(y2-y3)x2(y3-y1)x3(y1-y2)]

Neha Diwach 6 years, 10 months ago

For a right angle triangle 1/2×b×h For an equilateral triangle √3/4×side^2 According to chapter 7 of NCERT 1/2|x1^2(y2-y3)+x2^2(y3-y1)+x3^2(y1-y2)|

Sunny Raj 6 years, 10 months ago

1/2×base ×height

Ashwani Maurya 6 years, 10 months ago

1/2 base× height
  • 1 answers

Dia Khurana@1608 6 years, 10 months ago

See its quite simple 1/ a + 1/b + 1/x = 1/ a + b + x Do one thing bring 1/x to right side with 1/a+b+xand take the l.c.m of both the sides Then cross multiply it and u will get ur value... of A and B I hope u got ur answer...?
  • 4 answers

Jais Ravi 6 years, 10 months ago

Answer:- alpha+beta=b(for your question) And alpha+beta=b/a(for my question)

Jais Ravi 6 years, 10 months ago

Your question is wrong the eq. should be ax^2+bx+c=0 where a does not equal to 0

Khushi Singh 6 years, 10 months ago

alpha + beta=b

Gourav Sharma 6 years, 10 months ago

α+β .= b

  • 1 answers

Puja Sahoo? 6 years, 10 months ago

Its given in ncert, plz check there, as its too long to type here, hope yu can understand ?
  • 1 answers

Dia Khurana@1608 6 years, 10 months ago

Its there in rd sharma
  • 1 answers

Mohit Mohit 6 years, 10 months ago

A=30°

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