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CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 8 - MCQExams.com
CBSE
Class 8 Maths
Algebraic Expressions And Identities
Quiz 8
Simplify: $$\cfrac { { m }^{ 8 }{ p }^{ 7 }{ r }^{ 12 } }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{ } } \times { p }^{ 2 }{ r }^{ 3 }{ m }^{ 4 }$$
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$${m}^{5}{p}^{8}{r}^{3}$$
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$${m}^{9}{p}^{8}{r}^{7}$$
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$${m}^{9}{p}^{7}{r}^{4}$$
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$${m}^{9}{p}^{8}{r}^{6}$$
Explanation
$$\cfrac { { m }^{ 8 }{ p }^{ 7 }{ r }^{ 12 } }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{ } } \times { p }^{ 2 }{ r }^{ 3 }{ m }^{ 4 }$$
$$\Rightarrow$$
$$\cfrac { { m }^{ 12 }{ p }^{ 9 }{ r }^{ 1 5} }{ { m }^{ 3 }{ r }^{ 9 }{ p }^{1 } } $$
$$\Rightarrow$$
$$ m ^9 p ^8 r ^{ 6 } $$ (option D)
$$\left( 3-\sqrt { 7 } \right) \left( 3+\sqrt { 7 } \right) =$$?
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$$4$$
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$$2$$
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$$6$$
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$$8$$
Explanation
Given,
$$({3}-{\sqrt { 7 } })$$
$$({3}+{\sqrt { 7 } })$$.
We know, $$(x-y)(x+y)$$ $$={x}^{2}-{y}^{2}$$.
Thus,
$$({3}-{\sqrt { 7 } })$$
$$({3}+{\sqrt { 7 } })$$
$$={3}^{2}-{(\sqrt { 7 } )}^{2}$$
$$=9-7=2$$.
Therefore, option $$B$$ is correct.
Evaluate: $$-3z^2(-4x^2+10z^4-4xz^5)$$
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$$12x^2z^2-30z^6+12z^10$$
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$$12x^2z^2-30z^8+12z^10$$
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$$12x^4-30z^6+12z^7$$
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$$12x^2z^2-30z^6+12xz^7$$
Explanation
$$-3z^2(-4x^2+10z^4-4xz^5)$$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
$$=$$ $$-3z^2.(-4x^2)+-3z^2.10z^4-(-3z^2).4xz^5$$
$$=$$ $$12x^2z^2-30z^6+12xz^7$$
Using binomial identities, expand the following expression $$(t+3)^2$$
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$$t^2-6t-9$$
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$$t^2+6t-9$$
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$$t^2+6t+9$$
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$$t^2-6t+9$$
Explanation
We need to expand $$(t+3)^2$$
We know that binomial identities for $$(a+b)^2=a^2+2ab+b^2$$
So, $$a = t, b = 3$$
$$=$$ $$t^2+2\times 3t+3^2$$
$$=$$ $$t^2+6t+9$$
Find the value of $$(2x+5)(2x-5)$$.
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$$4x-25$$
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$$4x^2-25$$
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$$2x^2-25$$
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$$4x^2-5$$
Explanation
We know that binomial identities for $$(x+y)(x-y)=x^2-y^2$$
$$\therefore (2x+5)(2x-5) = (2x)^2-5^2$$
$$=$$ $$4x^2-25$$
Multiply: $$6x^5(-3x^4-2x^3+x^2)$$
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$$-18x^9-12x^8+6x^7$$
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$$-18x^5-12x^8+6x^7$$
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$$-18x^9-12x^5+6x^7$$
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$$-18x^9-12x^8+7x^7$$
Explanation
$$6x^5(-3x^4-2x^3+x^2)$$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
$$=$$ $$6x^5.-3x^4-6x^5.2x^3+6x^5.x^2$$
$$=$$ $$-18x^9-12x^8+6x^7$$
Simplify the following expression: $$(s-r)^2-(s+r)^2$$
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$$2sr$$
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$$2s^2+2r^2$$
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$$-4sr$$
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$$2s^2-2r^2$$
Explanation
We know that binomial identities for $$(x+y)^2=x^2+2xy+y^2$$
and $$(x-y)^2=x^2-2xy+y^2$$
So, $$(s-r)^2-(s+r)^2$$
$$=$$ $$s^2-2sr+r^2-s^2-2sr-r^2$$
$$=$$ $$-4sr$$
The value of $$-3x(5k^2+y)$$ is
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$$-9k^2x-3xy$$
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$$-15k^2x-3xy$$
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$$-15k^3x-3xy$$
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$$-5k^2x-3xy$$
Explanation
We need to find value of $$-3x(5k^2+y)$$
Multiply $$-3x$$ to the inside bracket, we get
=$$-15k^2x-3xy$$
Multiply: $$4x^2z(3x^2y^2-4xz)$$
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$$2x^4y^2z-16x^3z^2$$
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$$12x^4y^2z-16x^3z^2$$
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$$12x^4yz-16x^3z^2$$
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$$12x^4y^2z-16x^2z^2$$
Explanation
$$4x^2z(3x^2y^2-4xz)$$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
= $$4x^2z.3x^2y^2-4x^2z.4xz$$
= $$12x^4y^2z-16x^3z^2$$
Simplify: $$7a^3(-4a^4+6a^5)$$
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$$14a^7$$
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$$14a^8$$
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$$-28a^7+42a^8$$
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$$-28a^7+42a^{15}$$
Explanation
$$7a^3(-4a^4+6a^5)$$
Multiply each term of the parenthesis by the monomial keeping the addition or subtraction sign same.
$$=$$ $$7a^3.-4a^4+7a^3.6a^5$$
$$=$$ $$-28a^7+42a^8$$
Reduce the following expression using binomial identities: $$(3x+5)(3x+10)$$.
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$$9x^2+45x+50$$
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$$9x^2-45x+50$$
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$$9x^2+45x-50$$
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$$9x^2+45-50$$
Explanation
We know that binomial identities for $$(x+y)(x+z)=x^2+(y+z)x+yz$$
So, $$x = 3x, y = 5, z = 10$$
$$=$$ $$(3x)^2+(5+10)3x+5\times 10$$
$$=$$ $$9x^2+45x+50$$
What is the product of $$(x-2x)^2$$?
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$$2x^2$$
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$$x-2x^2$$
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$$x^2$$
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$$8x^2$$
Explanation
$$(x-2x)^2 = (x-2x)(x-2x)$$
$$=$$ $$x.x-x.2x-2x.x+2x.2x$$
$$=$$ $$x^2-2x^2-2x^2+4x^2$$
$$=x^2$$
If $$x\Box y=(x+y)^2-(x-y)^2$$. Then $$\sqrt 4\Box \sqrt 5=$$.
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$$0$$
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$$8\sqrt5$$
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$$10\sqrt5$$
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$$15$$
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$$20$$
The product of $$4{ a }^{ 2 }$$, $$-6{ b }^{ 2 }$$ and $$3{ a }^{ 2 }{ b }^{ 2 }$$ is
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$${ a }^{ 2 }{ b }^{ 2 }$$
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$$13{ a }^{ 4 }{ b }^{ 4 }$$
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$$-72{ a }^{ 4 }{ b }^{ 4 }$$
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$${ a }^{ 4 }{ b }^{ 4 }$$
Explanation
$$4a^{2}\times (-6b^{2})\times 3a^{2}b^{2}$$
$$=4\times a\times a\times -6\times b\times b\times 3\times a\times a\times b\times b$$
$$=4\times -6\times 3\times a\times a\times a\times a\times b\times b\times b\times b$$
$$=-72 a^{4}b^{4}$$
The value of the product $$\left( 4{ a }^{ 2 }+3b \right) \left( 9{ b }^{ 2 }+4a \right) $$ at $$a=1$$, $$b=-2$$ is
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$$60$$
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$$-80$$
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$$70$$
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$$-50$$
Explanation
$$(4a^{2}+3b)(9b^{2}+4a)$$
$$=36a^{2}b^{2}+27b^{3}+16a^{3}+12ab$$
Put the value of $$a=1$$ and $$b=-2$$, we get
$$(4a^{2}+3b)(9b^{2}+4a)\\$$
$$=36\times (1)^{2}\times (-2)^{2}+27\times (-2)^{3}+16\times (1)^{3}+12\times (1)\times (-2)\\$$
$$=36\times 4-27\times 8+16\times 1-12\times 2\\$$
$$=144-216+16-24=-80$$
The product of the monomials $$\displaystyle \left( 2{ x }^{ 4 }y \right)$$ and $$\left( 3{ x }^{ 5 }{ y }^{ 8 } \right) $$ is :
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$$\displaystyle 5{ x }^{ 9 }{ y }^{ 9 }$$
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$$\displaystyle 6{ x }^{ 9 }{ y }^{ 8 }$$
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$$\displaystyle 6{ x }^{ 9 }{ y }^{ 9 }$$
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$$\displaystyle 5{ x }^{ 20 }{ y }^{ 8 }$$
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$$\displaystyle 6{ x }^{ 20 }{ y }^{ 8 }$$
Explanation
The product of monomials $$2x^4y$$ and $$3x^5y^8$$ is
$$ 2x^4y\times 3x^5y^8= 2(3).x^4(x^5).y^1(y^8)$$
$$\Rightarrow 6x^{4+5}y^{1+8}$$
$$\Rightarrow 6x^9y^9$$
Number of terms in the expansion $$\left( a+b \right) \left( c+d \right) $$ is ________.
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
given, $$(a+b)(c+d)$$
$$=ac+bc+ad+bd$$
In above expression the number of terms are Four (4)
Multiply the binomials $$\displaystyle \left( 4z+3 \right) $$ and $$\left( z-2 \right) $$ :
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$$\displaystyle 4{ z }^{ 2 }-5$$
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$$\displaystyle 4{ z }^{ 2 }-6$$
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$$\displaystyle 4{ z }^{ 2 }-3z-5$$
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$$\displaystyle 4{ z }^{ 2 }-5z-6$$
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$$\displaystyle 4{ z }^{ 2 }+5z-6$$
Explanation
On multiplying, $$(4z+3)(z-2)$$
$$\Rightarrow 4z(z-2)+3(z-2)$$
$$\Rightarrow 4z^2-8z+3z-6$$
$$\Rightarrow 4z^2-5z-6$$
$$-4p\left( 3q-5p \right) =$$
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$$-12pq-20{ p }^{ 2 }$$
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$$-12pq+20{ p }^{ 2 }$$
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$$-7pq+20{ p }^{ 2 }$$
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$$-7pq-20{ p }^{ 2 }$$
Explanation
Multiplying $$-4p$$ with $$(3q-5p)$$
$$-4p(3q) = -12pq$$ $$-4p(-5p)$$
$$= -12pq+20p^2$$
What is the product $$(x + 2)(x - 3)$$?
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$$2x-6$$
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$$3x-2$$
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$$x^2-x-6$$
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$$x^2-6x$$
Explanation
$$(x+a)(x+b)$$ $$=$$$$x^2+(a+b)x+ab$$
In this formula, taking $$x=x$$ $$a= 2$$ $$b= (-3)$$
$$(x+2)(x-3)$$ = $$x^2+[(2+ (-3)]x+2(-3)$$
$$\Rightarrow$$ $$x^2-1x-6$$
$$\Rightarrow$$ $$\boxed{x^2-x-6}$$
The value of $$(7.2)^2$$ is (use an identity to expand):
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$$49.9$$
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$$14.4$$
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$$51.84$$
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$$49.04$$
Explanation
We will use, $$(a+b)^2$$= $$a^2+2ab+b^2$$ to find the required result.
Substitute $$7$$ for $$a$$ and $$0.2$$ for $$B$$ in the above identity,
$$(7+0.2)^2=(7)^2+2(7)(0.2)+(0.2)^2$$
$$=49+2.8+0.04$$
$$=51.84$$
Hence, option $$C$$ is correct.
The expansion of $$(2x - 3y)^2$$ is:
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$$2x^2+3y^2+6xy$$
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$$4x^2+9y^2-12xy$$
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$$2x^2+3y^2-6xy$$
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$$4x^2+9y^2+12xy$$
Explanation
Given,
$$(2x-3y)^2$$.
We know,
$$(a-b)^2$$ $$ =$$ $$a^2-2ab+b^2$$.
Then,
$$(2x-3y)^2$$ $$ =$$ $$(2x)^2-2 (2x)(3y)+(3y)^2$$
$$=$$ $$4x^2-12xy+9y^2$$.
Therefore, option $$B$$ is correct.
Which of the following relation is correct.
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$$3(x - 9) = 3x - 27$$
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$$3(x - 9) = 3x - 24$$
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$$3(x - 8) = 3x - 27$$
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None
Explanation
Let us multiply the given monomial $$3$$ by a binomial $$x-9$$ as shown below:
$$3(x-9)=(3\times x)-(3\times 9)=3x-27$$
Hence, $$3(x-9)=3x-27$$.
The product of the following monomial:
$$-3a,4a,b,-6c,d$$ is
$$72{a}^{2}bcd$$
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True
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False
Explanation
We multiply the given monomials $$-3a,4a,b,-6c$$ and $$d$$ as shown below:
$$(-3a)\times 4a\times b\times (-6c)\times d\\ =(-3\times a)\times 4a\times b\times (-6\times c)\times d\\ =(-3\times 4\times (-6))\times (a\times a)\times b\times c\times d\\ =72\times a^{ 2 }\times bcd\\ =72a^{ 2 }bcd$$
Hence, the product of
$$-3a,4a,b,-6c$$ and $$d$$
is $$72a^2bcd$$.
$$(a + b)^{2} - (a - b)^{2}=$$ _____.
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$$4ab$$
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$$2ab$$
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$$a^{2} + 2ab + b^{2}$$
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$$2(a^{2} + b^{2})$$
Explanation
Given, $$(a+b)^{2}-(a-b)^{2}$$.
We know,
$$(a+b)^{2}$$
$$=\left ( a^{2}+2ab+b^{2} \right )$$
and
$$(a-b)^{2}$$
$$=(a^{2}-2ab+b^{2})$$.
Then,
$$(a+b)^{2}-(a-b)^{2}$$
$$=\left ( a^{2}+2ab+b^{2} \right )-(a^{2}-2ab+b^{2})$$
$$=a^{2}+2ab+b^{2}-a^{2}+2ab-b^{2}\\=4ab$$.
Therefore, the required answer is $$4ab$$.
Hence, option $$A$$ is correct.
If $$A=xy, B=yz$$ and $$C=zx$$ then find ABC=
$${x}^{2}{y}^{2}{z}^{2}$$
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True
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False
Explanation
We multiply the given binomials $$A=xy,B=yz$$ and $$C=zx$$ as shown below:
$$(xy)\times (yz)\times (zx)\\ =(x\times y)\times (y\times z)\times (z\times x)\\ =(x\times x)\times (y\times y)\times (z\times z)\\ =x^{ 2 }\times y^{ 2 }\times z^{ 2 }\\ =x^{ 2 }y^{ 2 }z^{ 2 }$$
Hence, the product of
$$A=xy,B=yz$$ and $$C=zx$$
is $$x^2y^2z^2$$.
On multiplying the binomials:
$$2a-9$$ and $$3a+4$$ , we get
$$6{a}^{2}-19a-36$$
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0%
True
0%
False
Explanation
We multiply the given binomials
$$(2a-9)$$ and $$(3a+4)$$
as shown below:
$$(2a-9)(3a+4)\\ =2a(3a+4)-9(3a+4)\\ =6a^{ 2 }+8a-27a-36\\ =6a^{ 2 }+(8a-27a)-36\quad \quad \quad \quad \quad \quad \left( \text{Combining like terms} \right) \\ =6a^{ 2 }+a(8-27)-36\\ =6a^{ 2 }-19a-36$$
Hence,
$$(2a-9)(3a+4)=6a^{ 2 }-19a-36$$
.
So, the given statement is true.
State whether the statement is True or False.
Product of $$(kl+lm)$$ and $$(k-l)$$ is equal to
$${k}^{2}l-k{l}^{2}+{l}^{2}m-klm$$
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0%
True
0%
False
Explanation
Product of $$(kl+lm)$$ and $$(k−l)$$ is ,
$$(kl+lm)\times (k−l)=kl(k-l)+lm(k-l)\\=k^2l-kl^2+klm-l^2m$$
Clearly $$k^2l-kl^2+klm-l^2m\neq k^2l-kl^2-klm+l^2m$$
Hence, The given statement is false.
Find the product of the following pairs:
$$-3l,-2m$$
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$$6lm$$
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$$-6lm$$
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$$4lm$$
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None of these
Explanation
We multiply the given monomials $$-3l$$ and $$-2m$$ as shown below:
$$(-3l)\times (-2m)\\ =(-3\times -2)\times (l\times m)\\ =6\times lm\\ =6lm$$
Hence, the product of
$$-3l$$ and $$-2m$$
is $$6lm$$.
Find the product of the following pairs: $$6,7k$$
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$$42k$$
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$$21k$$
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$$23k$$
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None of these
Explanation
We multiply the given monomials $$6$$ and $$7k$$ as shown below:
$$\Rightarrow 6\times 7k\\ =(6\times 7)\times k\\ =42\times k\\ =42k$$
Hence, the product of
$$6$$ and $$7k$$ is $$42k$$.
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