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Algebraic Expressions And Identities Lessons
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Post a LessonAnswered on 02 Feb Learn CBSE/Class 8/Maths/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 CBSE/Class 8/Maths/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 CBSE/Class 8/Maths/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.
Answered on 02 Feb Learn CBSE/Class 8/Maths/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 CBSE/Class 8/Maths/Algebraic Expressions And Identities
Pooja R. Jain
To evaluate the product 8.56×11.608.56×11.60, you can simply multiply the two numbers:
8.56×11.60=99.5368.56×11.60=99.536
So, 8.56×11.608.56×11.60 is equal to 99.53699.536.
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