How to multiply variables - ISEE Middle Level Math

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Question

$6^{3} \div (11-5)=$

Answer

First find the exponent value: $6^{3}=216$

Then find the value of $\left ( 11-5 \right )=6$

Finally, solve the entire expression with the known values: $216\div 6=36$

The answer is 36.

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Question

Multiply:

Answer

Use the distributive property:

$4x(3x-2)=(4x\times3x)-(4x\times2)=12x^2-8x$

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Question

Simplify:

Answer

$=4 \cdot 2x + 4 \cdot 7 - 4$

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Question

$7n\cdot 3n=$

Answer

Multiply the numbers and multiply the variables:

$7n\cdot 3n=21n^{2}$

Answer: $21n^{2}$

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Question

$14s \cdot 3s=$

Answer

Multiply the numbers and multiply the variables:

$14\cdot 3\cdot s\cdot s=42s^{2}$

Answer: $42s^{2}$

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Question

Mulitply:

Answer

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Question

Simplify:

$(4v^{3})(xy^{2}v)(2x^{3}v)$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(4 * 2) (v^{3}vv)(xx^{3})(y^{2})$

When multiplying variables of the same type, you add* their exponents together. This gets you:

$(8)(v^{3+1+1})(x^{1+3})(y^2)$

$(8) (v^{5})(x^{4})(y^{2})$

This is the same as:

$8x^{4}y^{2}v^{5}$

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Question

Simplify:

$e * t * f^{4} * t^{2} * 10 * f$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(10)(e)(f^{4} f)(t t^{2})$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(10)(e)(f^{4+1})(t^{1+2})$

$(10)(e)(f^{5})(t^{3})$

This is the same as:

$10ef^{5}t^{3}$

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Question

Simplify:

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

When multiplying variables of the same type, you add their exponents together. (All of these variables have exponents of 1.) This gets you:

$(12)(e^{1+1+1+1+1+1+1})(f^{1+1+1+1+1+1})$

$(12)(e^{7}) (f^{6})$

This is the same as:

$12e^{7}f^{6}$

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Question

Simplify:

$2(f^{6})(u^{7})(e^{2}f^{2})(eu)$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(2)(e * e^{2})(f^{6}f^{2})(uu^{7})$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(2)(e^{1+2})(f^{6+2})(u^{1+7})$

$(2)(e^{3})(f^{8})(u^{8})$

This is the same as:

$2e^{3}f^{8}u^{8}$

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Question

Simplify:

$(2a)(3r^{2})(3a^{3})(5a^{3})(2a^{3})(b^{2})$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(2 3 3 5 2)(a a^{3} a^{3} * a^{3})(b^{2})(r^{2})$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(180)(a^{1+3+3+3})(b^2)(r^2)$

$(180)(a^{10})(b^{2})(r^{2})$

This is the same as:

$180a^{10}b^{2}r^{2}$

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Question

Simplify:

$(4g)(2r^{2})(r^{3})(g^{2})$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(24)(gg^{2})(r^{2}r^{3})$

When multiplying variables of the same type, you add* their exponents together. This gets you:

$(8)(g^{1+2})(r^{2+3})$

$(8)(g^{3})(r^{5})$

This is the same as:

$8g^{3}r^{5}$

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Question

Simplify:

$(2)(15y^{2})(x^{3})(10)$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(2 * 15 * 10)(x^{3})(y^{2})$

There are no variables to combine. We only need to combine the numerical coefficient. This gets you:

$(300)(x^{3})(y^{2})$

This is the same as:

$300x^{3}y^{2}$

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Question

Simplify:

$x * y * (x^{3}) * (xy)$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(x * x* x^{3})(y * y)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(x^{1+1+3})(y^{1+1})$

$(x^{5})(y^{2})$

This is the same as:

$x^{5}y^{2}$

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Question

Simplify:

$x * a^{3} * y^{4} * x^{2} * x*a^{2}$

Answer

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(a^{3} * a^{2})(x * x^{2} * x)(y^{4})$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(a^{3+2})(x^{1+2+1})(y^4)$

$(a^{5})(x^{4})(y^{4})$

This is the same as:

$a^{5}x^{4} y^{4}$

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Question

Simplify:

$xy(x^{2} + 2aby)$

Answer

$xy(x^{2} + 2aby)$

Distribute the outside term into the parentheses. Multiply each term in parentheses by :

$(xy)(x^{2}) + (xy)(2aby)$

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

$(xx^{2})(y) + (2)(a)(b)(x)(yy)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(x^{1+2})(y)+(2)(a)(b)(x)(y^{1+1})$

$(x^{3})(y) + (2)(a)(b)(x)(y^2)$

This is the same as:

$x^{3}y + 2abxy^{2}$

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Question

Simplify:

$-8 (-2y^{2}+ 7y - 9)$

Answer

$-8 (-2y^{2}+ 7y - 9)$

$= -8\cdot (-2y^{2} ) +\left ( -8 \right ) \cdot 7y -\left ( - 8\right ) \cdot 9$

$= 16y^{2} -56y+72$

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Question

Simplify:

Answer

Distribute the outside term into the parentheses. Multiply each term in parentheses by :

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

When multiplying variables of the same type, you add their exponents together. This gets you:

$(10)(abxy)-(6)(a^{1+1})(bx)+(8)(a)(b^{1+1})(y)$

$(10)(abxy) - (6)(a^{2})(bx)+ (2*4)(a)(b^{2})(y)$

This is the same as:

$10abxy - 6a^{2}bx + 8ab^{2}y$

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Question

Simplify:

$5bv(vx^{2} - 10bx - 5vb^{6})$

Answer

$5bv(vx^{2} - 10bx - 5vb^{6})$

Distribute the outside term into the parentheses. Multiply each term in parentheses by :

$(5bv)(vx^{2}) - (5bv)(10bx) - (5bv)(5vb^{6})$

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

$(5)(b)(v * v)(x^{2}) - (5 * 10)(b b)(vx) - (55)(b b^{6})(vv)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(5)(b)(v^{1+1})(x^2)-(50)(b^{1+1})(vx)-(25)(b^{1+6})(v^{1+1})$

$(5)(b)(v^{2})(x^{2}) - (50)(b^{2})(vx) - (25)(b^{7})(v^{2})$

This is the same as:

$5bv^{2}x^{2} - 50b^{2}vx - 25b^{7}v^{2}$

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Question

Multiply:

Answer

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