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Ask QuestionPosted by Nisha Murali 5 years, 1 month ago
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Posted by Nisha Murali 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
Total number of trials = 200
(i) P(getting 3 heads) = 
(ii) P(at least two heads) = 
(iii) P(two heads and one tail) = 
Posted by K. K. 5 years, 1 month ago
- 0 answers
Posted by Sana Mohammed Adam 5 years, 1 month ago
- 1 answers
Omveer Sehrawat 5 years, 1 month ago
Posted by Rachana Kumari 5 years, 1 month ago
- 4 answers
Omveer Sehrawat 5 years, 1 month ago
Posted by Manoj . 5 years, 1 month ago
- 3 answers
Kripa Daga 5 years, 1 month ago
Ibrahim Khalil 5 years, 1 month ago
Yogita Ingle 5 years, 1 month ago
Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°. The sides of a right triangle (say a, b and c) which have positive integer values, when squared, are put into an equation, also called a Pythagorean triple.
Consider the triangle given above:
Where “a” is the perpendicular side,
“b” is the base,
“c” is the hypotenuse side.
According to the definition, the Pythagoras Theorem formula is given as:
| Hypotenuse2 = Perpendicular2 + Base2
c2 = a2 + b2 |
Posted by Rohan King 5 years, 1 month ago
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Posted by Muskan Garg 5 years, 1 month ago
- 2 answers
Gaurav Seth 5 years, 1 month ago
It is given that ,
p = 2 - a -------( 1 )
LHS = a³ + 6ap + p³ - 8
= a³ + 6a( 2 - a ) + ( 2-a )³ - 8 [ from(1)]
= a³ +12a-6a²+2³-3×2²a+3×2a²-a³-8
= a³ + 12a - 6a² + 8 - 12a + 6a²- a³ - 8
= 0
= RHS
Hence proved.
Posted by Hardik Pandey 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
If we draw one diagonal to quadrilateral we will get two triangles i.e. △ABC and △ACD and we know that sum of the angles of a triangle is 180o
Sum of angles of quadrilateral ABCD=sum of angles of △ABC + sum of the angles △ACD
= 180o+180o
= 360o
Posted by P Nagaraju Nagaraju 5 years, 1 month ago
- 5 answers
Posted by Swatantra Yadav 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
Steps of Construction:
1.Draw a line segment PQ = 11 cm.( = AB + BC + CA).
2.At P construct an angle of 60° and at Q, an angle of 45°.
3.Bisect these angles. Let the bisectors of these angles intersect at a point A.
4.Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.
5.Join AB and AC (see Fig. 11.9). Then, ABC is the required triangle.
Posted by Komal Jaiswal 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
Sin30°=1/2
sin90°=1
cos0°=1
tan30°=1√3
tan60°=√3
now putting these these value on question
so..we find
=1/2-1+2×1/1/√3 ×√3
=1/2-1+2/√3/√3
=1-2+4/2
=3/2
Posted by Kaustuk Sharma 5 years, 1 month ago
- 5 answers
Posted by Yash Yash 5 years, 1 month ago
- 2 answers
Ashmita Das 5 years, 1 month ago
Yogita Ingle 5 years, 1 month ago
A pipe will have two layers as shown in picture.
There is a inner Cylinder and Outer Cylinder.
Inner Radius = Inner Diameter / 2 = 4 / 2 = 2 cm = r
Outer Radius = Outer Diameter / 2 = 4.4/2 = 2.2 cm = R
Height of the pipe = 77 cm = h
Inner Surface area of the Pipe = 2∏rh = 2×22×2×77/7 = 968 Square cm.
Outer Surface area of the pipe = 2∏Rh = 2×22×2.2×77/7 = 1064.8 Square cm.
Total Surface area = Inner Surface area + Outer Surface area + 2 × Cross section surface area.
The cross section suface area is present at two ends of the pipe.
This is the area covered between 2 circles.
Total Surface area = Inner Surface area + Outer Surface area + 2 × Cross section surface area.
= 968 + 1064.8 + 2×∏(R^2 – r^2)
= 2032.8 + 2×22×(2.2×2.2 – 2×2)/7
= 2032.8 + 5.28
= 2038.08 Square cm.
Posted by Vivek Singh 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
84-2n-2n²
= 2(42-n-n²) /* Taking 2 common */
= 2( 42 - 7n + 6n - n²) /* splitting the middle term */
= 2[7(6-n)+n(6-n)]
= 2(6-n)(7+n)
Therefore,.
2, (6-n) and (7+n) are factors of given quadratic expression.
Posted by Damini Damini 5 years, 1 month ago
- 1 answers
Posted by Anish Kumar 5 years, 1 month ago
- 2 answers
Shalini Kumari 5 years, 1 month ago
Posted by Damini Damini 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years, 1 month ago
Let r be the radius of a vessel.
Given: cost to paint the inner curved surface area= ₹2200
Cost to paint per m²= ₹20 & height of vessel (h)= 10 m
Curved surface area= 2πrh
Inner curved surface area=cost to paint the inner curved surface area/cost of paint per m²
2πrh= 2200/20= 110 m²
2× 22/7×r× 10= 110
r=(110×7)/(2×220)= 7/4= 1.75m
i)
Inner curved surface area of the vessel= 110 m²
ii)
Radius of the base radius of the base= 1.75 m
iii) capacity of The vessel= volume of the vessel
=πr²h= 22/7×7/4×7/4×10= 96.25 m³
= 96.25× 1000 L= 96.25 kL
[ 1 m= 100cm, 1 m³= 100 cm³
1000 cm³= 1L, 1000L= 1kL]
capacity of The vessel= 96.25 kL
Posted by Damini Damini 5 years, 1 month ago
- 1 answers
Posted by Damini Damini 5 years, 1 month ago
- 1 answers
Meenkashi Singh 5 years, 1 month ago
Posted by Neshok Ls 5 years, 1 month ago
- 3 answers
Simran Nehiniwal 5 years, 1 month ago
Simran Nehiniwal 5 years, 1 month ago
Gaurav Seth 5 years, 1 month ago
Given,
Lateral surface area = 94.2cm²
Height (h) = 5cm
Let the radius of the base be ' r '
We know,
Lateral surface area = 2πrh
94.2 = 2 × 3.14 × r × 5
94.2/ 2×3.14×5 = r
r = 3
Thus,
The radius of base is 3cm
Also,
Volume of the cylinder = πr²h
= 3.14 × 3² × 5 cm
= 3.14 × 9 × 5cm
= 141.3cm³
Thus,
The radius of the base is 3cm and the volume is 141.3cm³
Posted by Neshok Ls 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years, 1 month ago
AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Prove that EB = ED and AE = CE.
Given: AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E.
To Prove: EB = ED and AE = CE.
Construction: From O draw OP ⊥ AB and OQ ⊥ CD. Join OE.
Proof: ∵ AB = CD | Given
∴ OP = OQ
| ∵ Equal chords of a circle are equidistant from the centre
Now in right ∆s OPE and OQE,
Hyp. OE = Hyp. OE | Common
Side OP = Side OQ | Proved above
PE - PB = QE - QD
EB = ED Proved.
BE + AB = ED + CD 
AE = CE
Posted by Neshok Ls 5 years, 1 month ago
- 1 answers
Meenkashi Singh 5 years, 1 month ago
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Riya Sharma💜🤞 5 years, 1 month ago
2Thank You