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Algebraic Expressions and Identities Test - 32

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Algebraic Expressions and Identities Test - 32
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Weekly Quiz Competition
  • Question 1
    1 / -0
    $$-2x\times 10y\times -3z=$$
    Solution
    $$-2x\times 10y\times -3z= -2\times 10\times -3\times x\times y\times z$$
                                   $$=60xyz$$
  • Question 2
    1 / -0
    $$-8yz\times -2xy=$$
    Solution
    $$-8yz\times -2xy = (-8\times -2)\times yz\times xy$$
                            $$= 16\times y^2zx=16xy^2z$$
  • Question 3
    1 / -0
    $$2xy\times 7yz\times 9xz$$
    Solution
    $$2xy\times 7yz\times 9xz= (2\times 7\times 9)\times xy\times yx\times xz$$
                                  $$=126x^2y^2z^2$$
  • Question 4
    1 / -0
    Find the product of $$x^2yz\times xy^2z^3$$.
    Solution
    $$x^2yz\times xy^2z^3 = x^2\times y\times z\times x\times y^2\times z^3$$
            $$x^2\times x\times y\times y^2\times z\times z^3=x^3y^3z^4$$
  • Question 5
    1 / -0
    Find the product $$(a+b)(a-2b)$$, when $$a=1$$, $$ b=2$$.
    Solution
    $$=(a+b)(a-2b) =a\times a+b\times (-2b)+b\times a+a\times (-2b)$$
                 $$=a^2-2b^2-ab$$

    $$a=1, b=2$$
               $$=1-2\times 2^2-1\times 2$$
               $$=-9$$
  • Question 6
    1 / -0
    Find the product $$\dfrac{x}{2}\times (2x+4xy+6z)$$.
    Solution
    $$\dfrac{x}{2} \times(2x+4xy+6x)$$
    $$=\dfrac{x}{2}\times 2x+\dfrac{x}{2}\times 4xy+\dfrac{x}{2}\times 6z$$
    $$=x^2+2x^2y+3xz$$
  • Question 7
    1 / -0
    Find the product $$12m\times (n+2p+3m)$$.
    Solution
    $$12m(n+2p+3m)$$
    $$=12m\times n +12 m\times 2p+12m\times 3m$$
    $$= 12mn+24mp+36m^2$$
  • Question 8
    1 / -0
    Multiply $$\displaystyle \left((\frac{2}{5}x+y)(x+\frac{3}{5}y)\right)$$
    Solution
    $$((\dfrac{2}{5}x+y)(x+\dfrac{3}{5}y)$$

    =$$\dfrac{2}{5}x^2+\dfrac{6}{25}xy+xy+\dfrac{3}{5}x^2$$

    =$$\dfrac{2}{5}x^2+\dfrac{31}{25}xy+\dfrac{3}{5}y^2$$


  • Question 9
    1 / -0
    Find the product $$(x+4)(x^2+2x+1)$$
    Solution
    $$(x+4)(x^2+2x+1) = x\times(x^2+2x+1)+4(x^2+2x+1)$$
                                          $$=x^3+2x^2+x+4x^2+8x+4$$
                                          $$=x^3+6x^2+9x+4$$
  • Question 10
    1 / -0
    Which of the following is correct?
    Solution
    We know, $$(a-b)^2=a^2-2ab+b^2$$.

    Then,
    $$(x-y)^2$$
    $$=x^2-2(x)(y)+y^2$$ $$= x^2-2xy+y^2$$.

    To verify this, let us consider:
    $$(x-y)^2$$ $$=(x-y)$$$$(x-y)$$
    $$=x(x-y)-y(x-y)$$
    $$=x^2-xy-yx+y^2$$
    $$=x^2-xy-xy+y^2$$
    $$=x^2-2xy+y^2$$.

    Hence, $$(x-y)^2$$ $$= x^2-2xy+y^2$$

    Therefore, option $$B$$ is correct.
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