Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Sakshi Sharma 5 years, 2 months ago
- 1 answers
Posted by Varshini Varshini 5 years, 2 months ago
- 1 answers
Yogita Ingle 5 years, 2 months ago
Let x is the given expression. So,
We can split 11 as 6 + 5. So,
We can write the above expression as follows :
We know that,
So,
Taking root both sides :
So, we can say that the square root of the given expression is . Hence, this is the required solution.
Posted by Jaya Jaya 5 years, 2 months ago
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Ashmita Das 5 years, 2 months ago
Posted by Shivam Rana 5 years, 2 months ago
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Yogita Ingle 5 years, 2 months ago
In Maths, rational numbers are represented in p/q form where q is not equal to zero. It is also a type of real number.
12/17, 9/11 and 3/5 are positive rational numbers
-2/17, 9/-11 and -1/5 are negative rational numbers
Posted by Samiya Praveen 5 years, 2 months ago
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Alok Sahani 5 years, 2 months ago
Posted by Rijul Arora 5 years, 2 months ago
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Sreevidhya Muralidharan 5 years, 2 months ago
Posted by Manjeet Rattan 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
The diagonals of a square are equal and bisect each other at right angles.
Detailed proof
Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. To prove
that the diagonals of a square are equal and bisect each other at right angles, we have to
prove AC = BD, OA = OC, OB = OD, and ∠AOB = 90º.
In ΔABC and ΔDCB,
AB = DC (Sides of a square are equal to each other)
∠ABC = ∠DCB (All interior angles are of 90)
BC = CB (Common side)
So, ΔABC ≅ ΔDCB (By SAS congruency)
Hence, AC = DB (By CPCT)
Hence, the diagonals of a square are equal in length.
In ΔAOB and ΔCOD,
∠AOB = ∠COD (Vertically opposite angles)
∠ABO = ∠CDO (Alternate interior angles)
AB = CD (Sides of a square are always equal)
So, ΔAOB ≅ ΔCOD (By AAS congruence rule)
Hence, AO = CO and OB = OD (By CPCT)
Hence, the diagonals of a square bisect each other.
In ΔAOB and ΔCOB,
As we had proved that diagonals bisect each other, therefore,
AO = CO
AB = CB (Sides of a square are equal)
BO = BO (Common)
So, ΔAOB ≅ ΔCOB (By SSS congruency)
Hence, ∠AOB = ∠COB (By CPCT)
However, ∠AOB + ∠COB = 1800 (Linear pair)
2∠AOB = 1800
∠AOB = 900
Hence, the diagonals of a square bisect each other at right angles.
Posted by Dj Gaming 5 years, 2 months ago
- 3 answers
Gaurav Seth 5 years, 2 months ago
A n s w e r : 0
Degree of constant polynomial is always Zero
As it doesn’t have any variable, because in actual degree of the polynomial is nothing but the highest power of variable x in it.
Posted by Abhijeet Tiwari 5 years, 2 months ago
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Sreevidhya Muralidharan 5 years, 2 months ago
Posted by Suraj Yadav 5 years, 2 months ago
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Gaurav Seth 5 years, 2 months ago
SURFACE AREA OF A CUBOID:
Let us consider a cuboid of length = 1 units Breadth = b units and height = h units
Then we have :
(i) Total surface area of the cuboid =2(l * b + b * h + h * l) sq. units
(ii) Lateral surface area of the cuboid = [2 (1 + b)* h] sq. units
(iii) Area of four walls of a room = [2 (1 + b)* h] sq. units. = (Perimeter of the base * height) sq. units
(iv) Surface area of four walls and ceiling of a room = lateral surface area of the room + surface area of ceiling =2(1+b)*h+l*b
(v) Diagonal of the cuboid = √l2 + b2 + h2
SURFACE AREA OF A CUBE :
Consider a cube of edge a unit.
(i) The Total surface area of the cube = 6a2 sq. units
(ii) Lateral surface area of the cube = 4a2 sq. units.
(iii) The diagonal of the cube = √3 a units.
For more click on the given link:
<a href="https://ncerthelp.com/text.php?ques=1585+Notes+For+Volume+and+Surface+Area++Class+9+Formulas+Download+PDF+%0D%0A" ping="/url?sa=t&source=web&rct=j&url=https://ncerthelp.com/text.php%3Fques%3D1585%2BNotes%2BFor%2BVolume%2Band%2BSurface%2BArea%2B%2BClass%2B9%2BFormulas%2BDownload%2BPDF%2B%250D%250A&ved=2ahUKEwis0YCLkbvtAhVIzTgGHep_DekQFjAIegQICxAC" rel="noopener" target="_blank">Volume and Surface Area Class 9 Notes For Class 9 Formulas ...</a>
Posted by Chinki Chinki 5 years, 2 months ago
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Posted by Deepanshu Rohila 5 years, 2 months ago
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Aakhya Verma 5 years, 2 months ago
Posted by Ikra Malik 5 years, 2 months ago
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Posted by Suraj Yadav 5 years, 2 months ago
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Rohit Rathore 5 years, 2 months ago
Posted by Mitali Nigam 5 years, 2 months ago
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Yukti Verma 5 years, 2 months ago
Posted by Suraj Yadav 5 years, 2 months ago
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Suraj Yadav 5 years, 2 months ago
Posted by Pavan Kumar 5 years, 2 months ago
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Yogita Ingle 5 years, 2 months ago
Since the given solution is (2,14)
∴x=2 and y=14
Then, one equation is x+y=2+14=16
x+y=16
And, second equation is x−y=2−14=−12
x−y=−12
And, third equation is y=7x
7x−y=0
So, we can find infinite equations because through one point infinite lines can pass.
Posted by Pavan Kumar 5 years, 2 months ago
- 1 answers
Yogita Ingle 5 years, 2 months ago
4x²+9y²+16z²+12xy + 24yz +16xz
= (2x)²+(3y)²+(4z)²+2(2)(3)xy +2(3)(4)yz + 2(2x)(4z)
= (2x + 3y +4z)(2x + 3y + 4z)
= (2x + 3y + 4z)2
Posted by Tik Tok Videos K K 5 years, 2 months ago
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Suraj Yadav 5 years, 2 months ago
Posted by Parv Jain 5 years, 2 months ago
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Posted by Bhavneet Kaur 5 years, 2 months ago
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Yogita Ingle 5 years, 2 months ago
ax2 + bx + c = 0
9x2 + 30x + 28 =0
a = 9, b = 30, c = 28
Now b2 - 4ac = (30)2 - 4(9) (28)
= 900 - 1008
=-108
∴ b2 - 4ac< 0
∴ no real roots
Posted by Nisarg Chauhan 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
In the given image above.
AB = AD
and
DC = BC
Now,
in traingle ACD and ABC
AD= AB (given)
DC = BC (given)
AC = AC (common)
Therefore,
Triangle ADC is congruent to triangle ABC By SSA congurence creteria.
Therefore,
By C.P.C.T.
angle DAC = angle BAC
Now in Triangle AOD and triangle AOB
angle DAC = angle BAC (proved above)
AD = AB (given)
AO = AO (common)
Therefore,
both the given triangles are congruent by SAS creteria.
now,
By CPCT
DO = OB
Angle AOD = angle AOB
by using linear pair,
angle AOD + angle AOB = 180°
(as both the angles are equal )
2 angle AOD = 180°
angle AOD = 90°
Similarly,
angle AOB = 90°.
Using the same technique,
In triangle DOC and BOC.
AO= OC
angle DOC = angle BOC = 90°
Hence, proved.
Posted by Krishn Gamer 5 years, 2 months ago
- 2 answers
Gaurav Seth 5 years, 2 months ago
Step-by-step explanation:
Given : A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet.
To find :
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 7m×3.5m costsर 510.
Solution :
i) The dimension of the box is
Length l= 1.5 m
Breadth b= 1.25 m
Height h= 65 cm = 0.65 m
Area of the sheet required is
The area of the sheet required for making the box is 5.45 meter square.
ii) The cost of sheet for it, if a sheet measuring 7m×3.5m costsर 510.
i.e, area of sheet 24.5 meter square cost Rs.510
So, 1 meter square cost
Now, The cost of 5.45 meter square cost
Therefore, The cost of sheet is Rs. 113.447.
Sukriti Anupam 5 years, 2 months ago
Posted by Kasvi Shukvasiya 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
“If two sides of a triangle are unequal then the longer side has greater angle opposite to it.”
Given: A ΔABC in which AC > AB (say)
To prove: ∠ABC > ∠ACB
Construction: Mark a point D on AC such that AB = AD.
Join BD.
Proof: In ΔABD
AB = AD (by construction)
⇒ ∠1 = ∠2 …(i) (angles opposite to equal sides are equal)
Now in ΔBCD
∠2 > ∠DCB (ext. angle is greater than one of the opposite interior angles)
⇒ ∠2 > ∠ACB …(ii) [∵ ∠ACB = ∠DCB]
From (i) and (ii), we get
∠1 > ∠ACB …(iii)
But ∠1 is a part of ∠ABC
∠ABC > ∠1 …..(iv)
Now from (iii) and (iv), we get
∠ABC > ∠ACB
Hence proved.
Posted by Nitin Kumar Jain 5 years, 2 months ago
- 1 answers
Posted by Priyansh Yadav 5 years, 2 months ago
- 3 answers
Devil Ninja Gurpanish 5 years, 2 months ago
Himangshu Das 5 years, 2 months ago
Yogita Ingle 5 years, 2 months ago
Let the four angles of the quadrilateral be 3x, 5x, 9x and 13x
3x+5x+9x+13x=360∘ [sum of all angles of the quadrilateral is 360∘]
30x=360∘
x=12∘
Hence, the angles of the quadrilateral are 3×12∘=36∘, 5×12∘=60∘,9×12∘=108∘and13×12∘=156∘
Posted by Piya Yadav 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
Find the amount of water displaced by a solid spherical ball of diameter 0.21m
A n s w e r :
Radius (r) of ball = (0.21/2) m = 0.105 m
Therefore, the volume of the sphere is 0.004851 m3.
Posted by Piya Yadav 5 years, 2 months ago
- 1 answers
Yogita Ingle 5 years, 2 months ago
Given: Diameter of the spherical ball (d)= 28 cm
Radius =d/2= 28/2 cm = 14 cm
Amount of water displaced by a solid spherical ball = Volume of spherical ball = 4/3πr³
= (4/3 × 22/7 × 14 × 14 × 14)
= 34496/3 = 11498.67 cm³
Amount of water displaced by a solid spherical ball= 11498.67 cm³
Posted by Pavan I 5 years, 2 months ago
- 3 answers
Rohit Rathore 5 years, 2 months ago
Queen 14 5 years, 2 months ago
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Yogita Ingle 5 years, 2 months ago
If the perimeter of a equatleral triangle area is 360 =3 (side)
side = 360/3 = 120
Area of equalteral triangle = √ 3/4 (side)= √3/4 * 120 = 30√ 3
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