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  • 1 answers

Mutyala Gayathri 6 years, 11 months ago

Take a triangle ABC with CD as parallel points on the side AB and AC. We need to prove CD||BC. By the bpt theorem we came to know that AC/CB = AD/DC let us assume on contradiction that cd is not ll to bc. If CD is not ll to BC then there must be another line which is ll to it. Let us assume that ll line as CE from C. If CE is ll to BC then AC/CB = AE/EC. On substituting bpt theorem and this we get AE/EC = AD/DC. Therefore, CE and CD is same line. Hence proved
  • 1 answers

Suhana Parvin 6 years, 11 months ago

(1by sin theta-sin theta) (1by cos theta- cos) (sin by cos +cos by sin) =(1-sin square theta/sintheta) (1-cos square theta/cos theta)[ (sin sq. +cos sq theta) /sin cos] =(cos sq. /sin) (sin sq. Theta/cos theta) (1/sin cos) =1
  • 2 answers

Mutyala Gayathri 6 years, 11 months ago

If there is any other way of sending pic i will send u the proof. Kk

Mutyala Gayathri 6 years, 11 months ago

Statement: if a line is drwn ll to 1 side of a triangle to intersect the other 2 sides in distinct points the other 2 sides are divided in the same ratio.
  • 2 answers

Rana Jha 6 years, 11 months ago

It's right

Rishabh Jain 6 years, 11 months ago

Pls check your ques
  • 4 answers

Suhana Parvin 6 years, 11 months ago

I agree with rashi

Puja Sahoo 6 years, 11 months ago

Right

Rashi Kashyap 6 years, 11 months ago

Jazzy aap plz ek baar check kar lo ki question thik hai ya nhi,may be question incomplete hai.

Ananya Sharma 6 years, 11 months ago

It has no answer because in this equation two variable
  • 3 answers

Rashi Kashyap 6 years, 11 months ago

According to Euclid Division Lemma, a=bq+r,b<r<0 a1=3q+0 a2=3q+1 a3=3q+2 a1=3q. (Cubing both side) (a1)3= (3q)3 (a1)3=3q×3q×3q (a1)3=(9)3q3 Let 3q3=m (a1)3=9m (a2)3=3q+1. (Cubing both side) (a2)3=(3q+1)3 (a2)3=(3q)3+2×(3q)2×1+2×3q×(1)2+(1)3 (a2)3=9(3q3+2×(3q)2+6q)+1 let 3q3+2×(3q)2+6q=m (a2)3=9m+1 a3=3q+2. (Cubing both side) (a3)3=(3q+2)3 (a3)3=(3q)3+2×(3q)2×2+2×3q×(2)2+(2)3 (a3)3=9(3q3+2×(3q)2×2+2×3q×(2)2)+8 Let3q3+2×(3q)2×2+2×3q×4=m (a3)3=9m+8 Hence proved

Yogita Ingle 6 years, 11 months ago

<div> <div> <div>

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.

Case II : When x = 3q + 1

then, x3 = (3q + 1)3

= 27q3 + 27q2 + 9q + 1

= 9 q (3q2 + 3q + 1) + 1

= 9m + 1, where m = q (3q2 + 3q + 1)

Case III. When x = 3q + 2

then, x3 = (3q + 2)3

= 27 q3 + 54q2 + 36q + 8

= 9q (3q2 + 6q + 4) + 8

= 9 m + 8, where m = q (3q2 + 6q + 4)

Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8.

</div> </div> </div>

Sujal Gupta 6 years, 11 months ago

Bshshshghdh
  • 4 answers

Vivek Sarma 6 years, 11 months ago

Ya the answer is correct. ( cos square theta).

Ananya Sharma 6 years, 11 months ago

Hi

Md Soyeb 6 years, 11 months ago

=(1-sinΦ)(1+sinΦ) =1-sin²Φ =Cos²Φ

Chētnà Pandey✌️ 6 years, 11 months ago

What's the question bro I can't understand??
  • 4 answers

Mahak Yadav 6 years, 11 months ago

Sum of all sides of triangle

Yogita Ingle 6 years, 11 months ago

Perimeter of polygon = addition of all sides
Perimeter of triangle = Adding all 3 sides
Perimeter of equilateral triangle=  3 × side

Chētnà Pandey✌️ 6 years, 11 months ago

AB + BC + CA (add all three sides of ∆)

Sapna Jain 6 years, 11 months ago

A+b+c
  • 1 answers

Medhavi Mishra 6 years, 11 months ago

Let the positive number be x A. T. Q., x/88=8 x=88*8=704 Then, x/11=704/11 quotient =64 and remainder =0
  • 2 answers

Prashant Chaudhary 6 years, 11 months ago

Nice

Yogita Ingle 6 years, 11 months ago

px2 + 3x + q = 0
x = 1
p + 3 + q = 0
p + q = -3 .......... (i)
x = -2
p(-2)2 + 3(-2) + q = 0
4p - 6 + q = 0
4p + q = 6 ........... (ii)
Subtract (i) and (ii)
3p = 9
p = 9/3 = 3
Put it in (i)
3 + q = -3
q = -3 -3
q = -6

  • 1 answers

Gaurav Seth 6 years, 11 months ago

Let the value of certificate purchased by Sarah in the 1st year be Rs x.

Value of certificate purchase by Sarah in the 2nd year = Rs x + Rs 25 = Rs (x + 25)

Value of certificate purchased by Sarah in the 3rd year = Rs (x + 25) + Rs 25 = Rs (x + 50)  and so on.

xx + 25, x + 50,...... are in A.P.

First term, a = x

Common difference, = 25

Given, Total value of certificates purchased by her after 20 years = Rs 7250

∴ Value of certificate purchased by her in the 1st year = Rs 125

Value of certificate purchased by her in the 10th year, a10

∴ Value of certificate purchased by her in the 10th year = Rs 350

  • 2 answers

Chetna Saini 6 years, 10 months ago

Please reply me

Gaurav Seth 6 years, 11 months ago

a+2d = 4 …(1)

a+8d = -8 …(2)

Deduct (1) from (2), to get

6d = -12, or d = -2.

From (1), a = 4–2d = 4+4 = 8.

So the numbers of the AP are- 8, 6, 4, 2, 0, -2, -4, -6, -8.

So the 5th term is zero.

  • 2 answers

Chetna Saini 6 years, 11 months ago

Sin 18°upon cos 72° Sin changes to cos So ,sin(90°-18°) upon cos 72° Sin will change to cos 72 °upon cos72° Then cos 72° will cancel with cos72° The answer will be (1)

Pooja Nangliya 6 years, 11 months ago

Answer is zero (0)
  • 1 answers

Yogita Ingle 6 years, 11 months ago

The segment of  a circle divides it into two region namely major segment and minor segment. The segment having larger area is known as the major segment and the segment having smaller area is known as minor segment.
Area of minor segment = θ/360 πr2
<nobr></nobr>Area of major segment = area of minor segment <nobr>–Area of triangle </nobr>

8.2
  • 2 answers

Sapna Jain 6 years, 11 months ago

Which question

Prashant Chaudhary 6 years, 11 months ago

Aapka question kya hai Bhai ?
  • 1 answers

Sapna Jain 6 years, 11 months ago

Formula is xsquare -(alpha +beeta)x +alpha beeta
  • 4 answers

Amisha Sharma ? 6 years, 11 months ago

1 year

Prashant Chaudhary 6 years, 11 months ago

Koi bata Sakta hai mujhe ki kisko kitne din hue iss app ko use Karte

Prashant Chaudhary 6 years, 11 months ago

Sab important questions hai Sab padho tabhi sahi hoga

Sapna Jain 6 years, 11 months ago

Full chapter is very important
  • 1 answers

Yogita Ingle 6 years, 11 months ago

we have that:
AB = AC
so
AB² = AC²
then:
(a-8)² + (2 + 2)² = (a-2)² + (2+2)²
a²- 16a + 64 +16 = a²- 4a + 4 +16
-16a + 64 = -4a + 4
-16a + 4a =4 - 64
-12a = -60;
12a = 60;
a = 60/12 = 5;
a = 5

The value of a is 5.

 

  • 1 answers

Suhana Parvin 6 years, 11 months ago

Ur question may be wrong
  • 2 answers

Maria Anna Alwin 6 years, 11 months ago

How is that the answer??
The options are
17m 31m 12m 24m

Syed Mohd. Najaf 6 years, 11 months ago

25✓2/ 2

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