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Lesson Posted on 05 Aug Learn Algebraic Expressions And Identities
Class-8-Maths L-5 Square & Square roots Ex 5.3 Q5 (c)
Ashish Bhatia
I am a teacher, having 15+ years of experience teaching in CBSE schools. I have taught students of classes...
Lesson Posted on 07 Jun Learn Algebraic Expressions And Identities
Class 8- Maths L-8 Algebraic Expressions & Identities
Ashish Bhatia
I am a teacher, having 15+ years of experience teaching in CBSE schools. I have taught students of classes...
Lesson Posted on 07 Jun Learn Algebraic Expressions And Identities
Class 8- Maths L-8 Algebraic Expressions & Identities
Ashish Bhatia
I am a teacher, having 15+ years of experience teaching in CBSE schools. I have taught students of classes...
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Answered on 02 Feb Learn Algebraic Expressions And Identities
Pooja R. Jain
To find the value of x2−15x2−51 at x=−1x=−1, substitute x=−1x=−1 into the expression:
(−1)2−15(−1)2−51
1−151−51
To combine the terms with a common denominator, express 1 as 5555:
55−1555−51
4554
So, the value of x2−15x2−51 at x=−1x=−1 is 4554.
Answered on 02 Feb Learn Algebraic Expressions And Identities
Pooja R. Jain
To find the value of x2+y2−10x2+y2−10 at x=0x=0 and y=0y=0, substitute these values into the expression:
(0)2+(0)2−10(0)2+(0)2−10
0+0−100+0−10
−10−10
So, the value of x2+y2−10x2+y2−10 at x=0x=0 and y=0y=0 is −10−10.
Answered on 02 Feb Learn Algebraic Expressions And Identities
Pooja R. Jain
To find the product of 9a9a, 4ab4ab, and −2a−2a, simply multiply the coefficients (numbers) and the variables together:
(9a)×(4ab)×(−2a)(9a)×(4ab)×(−2a)
First, multiply the coefficients:
9×4×(−2)=−729×4×(−2)=−72
Now, multiply the variables:
a×a×b×a=a3ba×a×b×a=a3b
Combine the results:
(−72)×a3b(−72)×a3b
So, the product of 9a9a, 4ab4ab, and −2a−2a is −72a3b−72a3b.
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Answered on 02 Feb Learn Algebraic Expressions And Identities
Pooja R. Jain
To simplify the expression (a+b+c)(a+b−c)(a+b+c)(a+b−c), you can use the distributive property (FOIL method):
(a+b+c)(a+b−c)(a+b+c)(a+b−c)
=a(a+b−c)+b(a+b−c)+c(a+b−c)=a(a+b−c)+b(a+b−c)+c(a+b−c)
Now, distribute each term:
=a2+ab−ac+ba+b2−bc+ca+cb−c2=a2+ab−ac+ba+b2−bc+ca+cb−c2
Combine like terms:
=a2+2ab−ac+b2−bc−c2=a2+2ab−ac+b2−bc−c2
So, the simplified form of (a+b+c)(a+b−c)(a+b+c)(a+b−c) is a2+2ab−ac+b2−bc−c2a2+2ab−ac+b2−bc−c2.
Answered on 02 Feb Learn Algebraic Expressions And Identities
Pooja R. Jain
To evaluate (95)2(95)2 using identities, you can use the square of a binomial formula:
(a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2
In this case, a=90a=90 and b=5b=5. Apply the formula:
(95)2=(90+5)2(95)2=(90+5)2
=902+2×90×5+52=902+2×90×5+52
=8100+900+25=8100+900+25
=9025=9025
So, (95)2(95)2 is equal to
Answered on 02 Feb Learn Algebraic Expressions And Identities
Pooja R. Jain
To calculate the volume of a cuboidal box, you multiply its three dimensions (length, width, and height). In this case, the dimensions are given as 5x5x, 3x23x2, and 7x47x4:
Volume=Length×Width×HeightVolume=Length×Width×Height
Volume=(5x)×(3x2)×(7x4)Volume=(5x)×(3x2)×(7x4)
Now, multiply the terms:
Volume=105x1+2+4Volume=105x1+2+4
Combine the exponents:
Volume=105x7Volume=105x7
So, the volume of the cuboidal box is 105x7105x7.
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Answered on 02 Feb Learn Algebraic Expressions And Identities
Pooja R. Jain
A histogram is used to represent the distribution of continuous data, typically grouped into intervals or bins. Let's analyze the given scenarios:
(i) The number of letters for different areas in a postman’s bag:
(ii) The height of competitors in an athletics meet:
Therefore, for the height of competitors in an athletics meet, a histogram would be more appropriate to show the distribution of heights.
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