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Ask QuestionPosted by Poorni Gowda 3 years, 10 months ago
- 1 answers
Posted by Vijayalaxmi Malagitti 3 years, 10 months ago
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Aakhya Verma 3 years, 10 months ago
Posted by Divyam Gupta 3 years, 10 months ago
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Aakhya Verma 3 years, 10 months ago
Posted by Harshit Rajput 3 years, 10 months ago
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Posted by Priyanka Sharma 3 years, 10 months ago
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Yogita Ingle 3 years, 10 months ago
5/√3-√5
Divide and multiply by √3+√5
= 5(√3+√5)/ (√3-√5)(√3+√5)
= 5(√3+√5)/ (√3)2 - (√5)2
= 5(√3+√5)/ 3 - 5
= = 5(√3+√5)/(-2)
- 5/2 (√3+√5)
Posted by Piya Shah 3 years, 10 months ago
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Posted by Sakshi Sharma 3 years, 10 months ago
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Yogita Ingle 3 years, 10 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
Posted by Varshini Varshini 3 years, 10 months ago
- 1 answers
Yogita Ingle 3 years, 10 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 3 years, 10 months ago
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Ashmita Das 3 years, 10 months ago
Posted by Shivam Rana 3 years, 10 months ago
- 3 answers
Yogita Ingle 3 years, 10 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 3 years, 10 months ago
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Alok Sahani 3 years, 10 months ago
Posted by Rijul Arora 3 years, 10 months ago
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Sreevidhya Muralidharan 3 years, 10 months ago
Posted by Manjeet Rattan 3 years, 10 months ago
- 2 answers
Gaurav Seth 3 years, 10 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 3 years, 10 months ago
- 3 answers
Gaurav Seth 3 years, 10 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 3 years, 10 months ago
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Sreevidhya Muralidharan 3 years, 10 months ago
Posted by Suraj Yadav 3 years, 10 months ago
- 3 answers
Gaurav Seth 3 years, 10 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 3 years, 10 months ago
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Aakhya Verma 3 years, 10 months ago
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Rohit Rathore 3 years, 10 months ago
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Yukti Verma 3 years, 10 months ago
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Suraj Yadav 3 years, 10 months ago
Posted by Pavan Kumar 3 years, 10 months ago
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Yogita Ingle 3 years, 10 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 3 years, 10 months ago
- 1 answers
Yogita Ingle 3 years, 10 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 3 years, 10 months ago
- 2 answers
Suraj Yadav 3 years, 10 months ago
Posted by Parv Jain 3 years, 10 months ago
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Posted by Bhavneet Kaur 3 years, 10 months ago
- 1 answers
Yogita Ingle 3 years, 10 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 3 years, 10 months ago
- 1 answers
Gaurav Seth 3 years, 10 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 3 years, 10 months ago
- 2 answers
Gaurav Seth 3 years, 10 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 3 years, 10 months ago
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Gaurav Seth 3 years, 10 months ago
Types of Triangles Based on Sides
According to the lengths of their sides, triangles can be classified into three types which are:
If all three sides have the same length then it is an EQUILATERAL triangle, if only two sides have the same length then it is an ISOSCELES triangle and if there are no sides that have the same length then it is a SCALENE triangle.
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