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CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 6 - MCQExams.com
CBSE
Class 7 Maths
Algebraic Expressions
Quiz 6
What should be taken away from $$3x^2-4y^2+5xy+20$$ to get $$ -x^2-y^2+6xy+20$$.
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$$x^2-y^2-xy$$
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$$x^3-3y^2-xy$$
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$$x^2-y^3-xy$$
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$$4x^2-3y^2-xy$$
Explanation
Let $$p$$ be the required expression. Then according the question, we have:
$$(3x^{ 2 }−4y^{ 2 }+5xy+20)-p=−x^{ 2 }−y^{ 2 }+6xy+20\\ \Rightarrow p=(3x^{ 2 }−4y^{ 2 }+5xy+20)-(-x^{ 2 }−y^{ 2 }+6xy+20)\\ \Rightarrow p=3x^{ 2 }−4y^{ 2 }+5xy+20+x^{ 2 }+y^{ 2 }-6xy-20\\ \Rightarrow p=(3x^{ 2 }+x^{ 2 })+(−4y^{ 2 }+y^{ 2 })+(5xy-6xy)+(20-20)\quad \quad \quad \quad (Combining\quad like\quad terms)\\ \Rightarrow p=4x^{ 2 }−3y^{ 2 }-xy$$
Hence,
$$(3x^{ 2 }−4y^{ 2 }+5xy+20)-(4x^{ 2 }−3y^{ 2 }-xy)=−x^{ 2 }−y^{ 2 }+6xy+20$$.
Find the like terms from the given terms and add them:
$$4xyz, 2x^2y^2, 3x^2y^2, 4x^2, 9x^2y^2, 18x^2, 6xyz, 10x^2, 7xyz$$
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(1) 11xyz
(2) $$12x^2y^2$$
(3) $$32x^2$$
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(1) 17xyz
(2) $$14x^2y^2$$
(3) $$32x^2$$
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(1) -17xyz
(2) $$14x^2y^2$$
(3) $$28x^2$$
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none
Explanation
We know that like terms are terms whose variables (and their exponents such as the $$2$$ in $$x^2$$) are the same.
In the given terms $$4xyz,2x^2y^2,3x^2y^2,4x^2,9x^2y^2,18x^2,6xyz,10x^2,7xyz$$, the like terms are as follows:
(i) $$4xyz,6xyz,7xyz$$
(ii) $$2x^2y^2,3x^2y^2,9x^2y^2$$
(iii) $$4x^2,18x^2,10x^2$$
Now, adding the like terms, we get:
(i) $$4xyz+6xyz+7xyz=17xyz$$
(ii) $$2x^2y^2+3x^2y^2+9x^2y^2=14x^2y^2$$
(iii) $$4x^2+18x^2+10x^2=32x^2$$
Hence, after adding the like terms we get $$17xyz, 14x^2y^2,32x^2$$.
State True or False
$$13m^2-2m^2 = 11m^2$$
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True
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False
Explanation
Let us solve the given expression $$13m^2-2m^2$$ by combining like terms as shown below:
$$13m^2-2m^2=(13-2)m^2=11m^2$$
Hence,
$$13m^2-2m^2=11m^2$$
.
Subtract $$4{ p }^{ 2 }q-3pq+5p{ q }^{ 2 }-8p+7q-10$$ from $$18-3p-11q+5pq-2{ q }^{ 2 }+5{ p }^{ 2 }q$$
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$$28+5p-18q+8pq-2q^2+p^2q-5pq^2$$
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$$8+11p-18q+8pq-2q^2+p^2q-5pq^2$$
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$$28-5p+18q-8pq+2q^2-p^2q+5pq^2$$
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none
Explanation
We subtract the given expression $$4p^2q-3pq+5pq^2-8p+7q-10$$ from $$18-3p-11q+5pq-2q^2+5p^2q$$ as shown below:
$$(18-3p-11q+5pq-2q^{ 2 }+5p^{ 2 }q)-(4p^{ 2 }q-3pq+5pq^{ 2 }-8p+7q-10)\\ =18-3p-11q+5pq-2q^{ 2 }+5p^{ 2 }q-4p^{ 2 }q+3pq-5pq^{ 2 }+8p-7q+10\\ =(18+10)+(-3p+8p)+(-11q-7q)+(5pq+3pq)-2q^{ 2 }+(5p^{ 2 }q-4p^{ 2 }q)-5pq^{ 2 }\\ (Combining\quad like\quad terms)\\ =28+5p-18q+8pq-2q^{ 2 }+p^{ 2 }q-5pq^{ 2 }$$
Hence,
$$(18-3p-11q+5pq-2q^{ 2 }+5p^{ 2 }q)-(4p^{ 2 }q-3pq+5pq^{ 2 }-8p+7q-10)=28+5p-18q+8pq-2q^{ 2 }+p^{ 2 }q-5pq^{ 2 }$$
.
Simplify: $$\left( { x }^{ 2 }-{ y }^{ 2 }+2xy+1 \right) -\left( { x }^{ 2 }-{ y }^{ 2 }+4xy-5 \right) $$
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$$-2xy+6$$
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$$2xy+6$$
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$$2x{ y }^{ 2 }-2xy+6$$
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$$-2{ y }^{ 2 }-2xy-6$$
Explanation
Given: $$(x^2-y^2+2xy+1)-(x^2-y^2+4xy-5)$$
$$=x^2-y^2+2xy+1-x^2+y^2-4xy+5$$
$$ = x^2-x^2-y^2+y^2+2xy-4xy+1+5$$
$$=-2xy+6$$
$$\therefore$$
$$(x^2-y^2+2xy+1)-(x^2-y^2+4xy-5)=-2xy+6$$
If A = $$10{w}^{3} + 20{w}^{2} - 55w + 60$$,
B = $$-25{w}^{2} + 15w - 10$$ and
C = $$5{w}^{2} - 10w + 20$$,
then A + B - C is equal to ______.
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$$10{w}^{3} + 10{w}^{2} + 30w + 30$$
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$$10{w}^{3} + 10{w}^{2} - 30w + 30$$
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$$10{w}^{3} - 10{w}^{2} - 30w + 30$$
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None of these
Explanation
So, $$A+B-C=(10W^3+20W^2-55W+60)+(-25W^2+15W-10)-(5W^2-10W+20)$$
Note- Polynomials with same order will be operated with each other.
So, $$A+B-C=10W^3+(20W^2-25W^2-5W^2)+(-55W+15W+10W)+(60-10-20)$$
So, $$A+B-C=10W^3-10W^2-30W+30$$
Simplify :
(3x + 2y - 9) (2x - 6y + 2) - [(4x - 9y - 1)
+ (-3x + 8y + 7)]
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$$6{x}^{2} - 14xy - 12{y}^{2} - 13x + 59y - 24$$
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$$6{x}^{2} - 12xy - 189 - 17x + 61y - 29$$
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$$8{x}^{2} - 14xy - 12{y}^{2} - 13x + 57y- 24$$
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$$8{x}^{2} - 14xy - 12{y}^{2} - 17x + 61y - 29$$
Explanation
Given expression is $$(3x+2y-9)(2x-6y+2)-[(4x-9y-1)+(-3x+8y+7)]$$
We can perform the given operations to simplify the expression further.
$$=(6x^2-18xy+6x+4xy+4y-12y^2-18x+54y-18)-[4x-9y-1-3x+8y+7]$$
$$\Rightarrow 6x^2-12y^2-14xy-12x+58y-18-x+y-6$$
$$\Rightarrow 6x^2-12y^2-14xy-13x+59y-24$$
Hence, $$(3x+2y-9)(2x-6y+2)-[(4x-9y-1)+(-3x+8y+7)]=6x^2-12y^2-14xy-13x+59y-24$$
By how much is $$a^{4} - 4a^{2}b^{2} + b^{4}$$ more than $$a^{4} - 8a^{2}b^{2} + b^{4}$$?
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$$4a^{2}b^{2}$$
0%
$$-12a^{2}b^{2}$$
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$$2a^{4} + 2b^{4}$$
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$$10a^{2}b^{2}$$
Explanation
We have two expressions
$$a^{4}-4a^{2}b^{2}+b^{2}$$ ............(1)
$$a^{4}-8a^{2}b^{2}+b^{2}$$ ............(2)
Subtract (2) from (1), we get
$$a^{4}-4a^{2}b^{2}+b^{2} - (a^{4}-8a^{2}b^{2}+b^{4})$$
$$= a^{4}-4a^{2}b^{2}+b^{2} - a^{4}+8a^{2}b^{2}-b^{4}$$
$$= -4a^{2}+8a^{2}b^{2}$$
$$= 4a^{2}b^{2}$$
What must be subtracted from $$3a^{2} - 6ab - 3b^{2} - 1$$ to get $$4a^{2} - 7ab - 4b^{2} + 1$$?
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$$a^{2} - ab - b^{2} + 2$$
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$$6a^{2} + 2ab + 8b^{2} - 7$$
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$$-a^{2} + ab + b^{2} - 2$$
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$$3a^{2} - ab + b^{2} - 2$$
Explanation
Let the polynomial to be subtracted be $$s$$. Then,
$$(3a^2-6ab-3b^2-1)-s=(4a^2-7ab-4b^2+1)$$
$$\Rightarrow s = \left( {3{a^2} - 6ab - 3{b^2} - 1} \right) - \left( {4{a^2} - 7ab - 4{b^2} + 1} \right)$$
$$\Rightarrow s = 3{a^2} - 6ab - 3{b^2} - 1 - 4{a^2} + 7ab + 4{b^2} -1 $$
$$\Rightarrow s = \left( {3{a^2} - 4{a^2}} \right) + \left( { - 6ab + 7ab} \right) + \left( { - 3{b^2} + 4{b^2}} \right) + \left( { - 1 - 1} \right)$$
$$\Rightarrow s = - {a^2} + ab - {b^2} - 2$$
Therefore, $$-a^2+ab+b^2-2$$ must be subtracted from $$3a^2-6ab-3b^2-1$$ to get $$4a^2-7ab-4b^2+1.$$
Hence, option $$C$$ is correct.
Add $$5x^2-7x+3, -8x^2+2x-5$$ and $$7x^2-x-2$$
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$$4x^2-7x+3$$
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$$4x^2-6x-4$$
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$$4x^2+7x-4$$
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$$4x^2-6x-3$$
Explanation
To simplify an algebraic expression that consists of both like and unlike terms, we need to move the like terms together and then add or subtract their coefficients.
Here, the terms are
$$5x^2-7x+3,-8x^2+2x-5$$ and $$7x^2-x-2$$. We add these terms as shown below:
$$(5x^{ 2 }-7x+3)+(-8x^{ 2 }+2x-5)+(7x^{ 2 }-x-2)\\ =\left( 5{ x }^{ 2 }-8{ x }^{ 2 }+7{ x }^{ 2 } \right) +\left( -7x+2x-x \right) +\left( 3-5-2 \right) \\ ={ x }^{ 2 }\left( 5-8+7 \right) +x\left( -7+2-1 \right) +\left( 3-5-2 \right) \\ =4{ x }^{ 2 }-6x-4$$
Hence,
$$(5x^{ 2 }-7x+3)+(-8x^{ 2 }+2x-5)+(7x^{ 2 }-x-2)=4{ x }^{ 2 }-6x-4$$.
Do the following subtractions as directed.
Subtract $$(8p+9k-17)$$ from $$(2pq+7p-8q+15)$$
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$$-p-8q-9k+2pq+32$$
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$$p+8q-9k+2pq+32$$
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$$-p-8q+9k+pq+32$$
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$$p-8q-9k+2pq-32$$
Explanation
We know that Like terms are terms that contain the same variables raised to the same power.
We group together the like terms and subtract $$8p+9k-17$$ from $$2pq+7p-8q+15$$ as follows:
$$(2pq+7p-8q+15)-(8p+9k-17)\\ =2pq+7p-8q+15-8p-9k+17\\ =(7p-8p)-8q-9k+2pq+(15+17)\\ =-p-8q-9k+2pq+32$$
Hence,
$$(2pq+7p-8q+15)-(8p+9k-17)=-p-8q-9k+2pq+32$$
.
Do the following subtractions as directed.
Subtract $$t^4-3t^2+7$$ from $$5t^3-9$$
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$$t^4-3t^2-5t^3+16$$
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$$-t^4-5t^3-3t^2+16$$
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$$-t^4+5t^3+3t^2-16$$
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$$t^4-5t^3-3t^2-16$$
Explanation
We know that Like terms are terms that contain the same variables raised to the same power.
The only like term in the given expressions $$t^4-3t^2+7$$ and $$5t^3-9$$ are $$7$$ and $$-9$$.
So we group together the like terms and subtract
$$t^4-3t^2+7$$ from
$$5t^3-9$$
as follows:
$$(5t^{ 3 }-9)-(t^{ 4 }-3t^{ 2 }+7)\\ =5t^{ 3 }-9-t^{ 4 }+3t^{ 2 }-7\\ =-t^{ 4 }+5t^{ 3 }+3t^{ 2 }+(-9-7)\\ =-t^{ 4 }+5t^{ 3 }+3t^{ 2 }-16$$
Hence, $$(5t^{ 3 }-9)-(t^{ 4 }-3t^{ 2 }+7)=-t^{ 4 }+5t^{ 3 }+3t^{ 2 }-16$$.
Do the following subtractions as directed.
Subtract $$(3x^3-7x+4)$$ from $$(5x^3+7x^2-2x)$$
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$$2x^3+4x^2+5x-2$$
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$$2x^3+7x^2+5x-2$$
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$$2x^3+7x^2+5x-4$$
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$$2x^3+x^2+5x-2$$
Explanation
We know that Like terms are terms that contain the same variables raised to the same power.
We group together the like terms and subtract $$3x^3-7x+4$$ from $$5x^3+7x^2-2x$$ as follows:
$$(5x^{ 3 }+7x^{ 2 }-2x)-(3x^{ 3 }-7x+4)\\ =5x^{ 3 }+7x^{ 2 }-2x-3x^{ 3 }+7x-4\\ =(5x^{ 3 }-3x^{ 3 })+7x^{ 2 }+(-2x+7x)-4\\ =2x^{ 3 }+7x^{ 2 }+5x-4$$
Hence,
$$(5x^{ 3 }+7x^{ 2 }-2x)-(3x^{ 3 }-7x+4)=2x^{ 3 }+7x^{ 2 }+5x-4$$
.
Add the following :
i) $$5y^3 , 26y^3, 10y^3, -3y^3$$
ii) $$3x^2, -10x^2, 4x^2$$
iii) $$4x^2y, -3xy^2, -5xy^2, 5x^2y$$
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(i) $$38y^3$$ (ii) $$-3x^2$$ (iii) $$8x^2y-9xy^2$$
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(i) $$38y^3$$ (ii) $$-3x^2$$ (iii) $$9x^2y-8xy^2$$
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(i) $$41y^3$$ (ii) $$-3x^2$$ (iii) $$9x^2y-8xy^2$$
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None
Explanation
i)
$$5y^3,26y^3,10y^3,−3y^3 $$
$$5y^3+26y^3+10y^3−3y^3 $$
$$y^3(5+26+10-3)$$
$$38y^3$$
ii)
$$3x^2,−10x^2,4x^2$$
$$3x^2−10x^2+4x^2$$
$$x^2(3-10+4)$$
$$-3x^2$$
iii)
$$4x^2y,−3xy^2,−5xy^2,5x^2y$$
$$4x^2y−3xy^2−5xy^2+5x^2y$$
$$4x^2y+5x^2y−3xy^2−5xy^2$$
$$9x^2y−8xy^2$$
Subtract $$2x^{3}-4x^{2}+3x+5$$ from $$4x^{2}+x^{2}+x+6$$, then the resultant value is
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$$6x^{3}+5x^{2}-2x+1$$
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$$2x^{3}+5x^{2}-2x+1$$
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$$2x^{3}-5x^{2}-2x+1$$
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$$2x^{3}-5x^{2}+2x-1$$
What must be added to $$5x^{3}-2x^{2}+6x+7$$ to make the sum $$x^{3}+3x^{2}-x+1$$?
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$$4x^{3}-5x^{2}+7x+6$$
0%
$$4x^{3}+5x^{2}-7x+6$$
0%
$$-4x^{3}+5x^{2}-7x-6$$
0%
$$None\ of\ these$$
Explanation
$$\left( {x}^{3} + 3 {x}^{2} - x + 1 \right) - \left( 5 {x}^{3} - 2 {x}^{2} + 6x + 7 \right)$$
$$= \left( 1 - 5 \right) {x}^{3} + \left( 3 + 2 \right) {x}^{2} - \left( 1 + 6 \right) x + 1 - 7$$
$$= -4 {x}^{3} + 5 {x}^{2} - 7x - 6$$
Hence $$\left( -4 {x}^{3} + 5 {x}^{2} - 7x - 6 \right)$$ must be added to $$\left( {x}^{3} + 3 {x}^{2} - x + 1 \right)$$ to make the sum $$\left( 5 {x}^{3} - 2 {x}^{2} + 6x + 7 \right)$$.
Simplify combining like terms :
$$p-(p-q)-q-(q-p)$$
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$$q-p$$
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$$q$$
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$$0$$
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$$p-q$$
Explanation
$$p-(p-q)-q-(q-p)$$
$$=p-p+q-q-q+p$$
$$=p-q$$
Add: $$2u+3v,\,\,-2v+3w,\,\,3u-v+2w$$
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$$5u-5w$$
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$$5u+5w$$
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$$u-w$$
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$$u+w$$
Explanation
given, $$\left(2u+3v+3u-v+2w\right)+\left(-2v+3w\right)$$
$$=5u+2v+2w-2v+3w$$
$$=5u+5w$$
What must be added to $${ 7z }^{ 3 }-{ 11z }^{ 2 }-129$$ to get $${ 5z }^{ 2 }+7z-92$$ ?
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$${ 7z }^{ 3 }+{ 16z }^{ 2 }+7z+37$$
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$$-{ 7z }^{ 3 }+{ 16z }^{ 2 }+7z+37$$
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$$-{ 7z }^{ 3 }+{ 16z }^{ 2 }+7z-37$$
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$$-{ 7z }^{ 3 }-{ 7z }^{ 2 }+7z-37$$
Explanation
Let a must be added.
Then,
$$7z^3 - 11z^2 -129 + a = 5z^2 +7z -92$$
$$\rightarrow a = -7z^3 + 16z^2 + 7z + 37$$
Thus, B is the correct answer.
Simplify: $$a-[b-\{c-(a-b-c)\}-c]+a$$
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$$a-2c$$
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$$a-3c$$
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$$a+3c$$
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$$a+2c$$
The sum of $$(6a+4b-c+3), (2b-3c+4), (11b-7a+2c-1)$$ and $$(2c-5a-6)$$ is
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$$(4a-6b+2)$$
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$$(-3a+14b-3c+2)$$
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$$(-6a+17b)$$
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$$(-6a+6b+c-4)$$
Explanation
$$(6a+4b−c+3)+(2b−3c+4)+(11b−7a+2c−1)+(2c−5a−6)$$
$$\Rightarrow$$
$$6a+4b−c+3+2b−3c+4+11b−7a+2c−1+2c−5a−6$$
Now arranging and combining the like terms,
$$\Rightarrow$$ $$(6a-7a-5a)+(4b+2b+11b)+(-c-3c+2c+2c)+(3+4-1-6)$$
$$\Rightarrow$$ $$-6a+17b$$
$$\therefore$$
The sum of $$(6a+4b−c+3),(2b−3c+4),(11b−7a+2c−1)$$ and $$(2c−5a−6)$$ is $$(-6a+17b)$$
The length of a side of square is given as $$2x+3$$. Which expression represents the perimeter of the square.
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$$2x+16$$
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$$6x+9$$
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$$8x+3$$
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$$8x+12$$
Explanation
Apply the formula of perimeter of square
Since, the perimeter of a square $$=side+side+side+side=4 \times side$$
The perimeter of the given square $$=4 \times (2x+3)$$
$$=4 \times 2 x+4 \times 3$$
$$\Rightarrow 4 \times(2 x+3)=8 x+12$$
Therefore, the perimeter of the given square $$=8x+12$$
$$\textbf{Hence, the correct option is (d)}$$
Number of terms in the expression $$3{ x }^{ 2 }{ y }-2{ y }^{ 2 }{ z }-{ z }^{ 2 }x+5$$ is
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$$2$$
0%
$$3$$
0%
$$4$$
0%
$$5$$
Explanation
The terms in the given expression are $$3 x^{2} y,-2 y^{2} z,-z^{2} x$$ and $$5$$
Therefore, the number of terms in the given expression are $$4 .$$
$$\textbf{Hence, the correct option is (c)}$$
The terms of the expression $$4{ x }^{ 2 }-3xy$$ are:
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$$4{ x }^{ 2 }$$ and $$-3xy$$
0%
$$4{ x }^{ 2 }$$ and $$3xy$$
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$$4{ x }^{ 2 }$$ and $$-xy$$
0%
$${ x }^{ 2 }$$ and $$xy$$
Explanation
We know that terms are added in an expression.
Therefore terms in the expression $$4 x^{2}-3 x y$$ are $$4 x^{2}$$ and $$-3 x y$$
$$\textbf{Hence, the correct option is (a)}$$
State true or false.
$$(3a-b+3)-(a+b)$$ is a binomial.
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True
0%
False
State true or false: $$7x$$ has two terms, $$7$$ and $$x$$
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0%
True
0%
False
Explanation
$$7x$$ is a multiplication of $$7$$ and $$x$$.
Since, $$7x$$ contains no addition and subtraction operations and therefore it has only one term
State the following statement is true or false.
Sum of $$x^2+x$$ and $$y^2+y$$ is $$2x^2+2y^2$$
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0%
True
0%
False
Explanation
$$x^2+x+y^2+y=x^2+y^2++y$$
Hence, s
um of $$x^2+x$$ and $$y^2+y$$ is not $$2x^2+2y^2$$
State the following statement is true or false.
If we subtract a monomial from a binomial, then the answer is at least a binomial.
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0%
True
0%
False
Explanation
Ans :- False
It can be a monomial
For example: 6x + (4 - 6x) = 4 which is monomial.
If the perimeter of rectangle is given by $$2(a+b)$$, where $$a$$ and $$b$$ denotes the length and breadth of the rectangle respectively.. The variable used here are ____ and ____.
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$$2$$ and $$a$$
0%
$$2$$ and $$b$$
0%
$$a $$ and $$b$$
0%
$$2, a$$ and $$b$$
If $$a$$ represent the side of a square, then perimeter of the square is ____.
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0%
$$2a$$
0%
$$4a$$
0%
$$6a$$
0%
$$8a$$
Explanation
Given side of square: $$a$$ units
We know $$\textit{Perimeter of square} = 4 \times side$$
$$\therefore$$ the required expression here is $$4a$$
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