How to multiply variables - ISEE Middle Level Quantitative Reasoning

Card 0 of 860

Question

Mulitply:

Answer

The FOIL method works nicely here:

$(x-7 )(x+5) = x \cdot x + x \cdot5 + (-7) \cdot x + (-7) \cdot 5 = x^{2} +5x -7x -35 = x^{2} - 2x -35$

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Question

Which quantity is greater?

$(a)\ (-4)^{-2}$

$(b)\ (-4)^{2}$

Answer

$(a) =(-4)^{-2} = \frac{1}{(-4)^{2}}=\frac{1}{(-4)(-4)}=\frac{1}{16}$

$(b)= (-4)^{2}=(-4)(-4)=16$

is the greater amount.

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Question

Which quantity is greater?

$(a)\ (2)^{-3}$

$(b)\ (-2)^{3}$

Answer

$(a) =(2)^{-3} = \frac{1}{(2)^{3}}=\frac{1}{8}$

$(b)= (-2)^{3}=(-2)(-2)(-2)=-8$

is the greater amount.

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Question

Which quantity is greater?

$(a)\ (-\frac{1}{2})^{4}$

$(b)\ (-\frac{1}{2})^{-4}$

Answer

$(a)= (-\frac{1}{2})^{4} = (-\frac{1}{2})(-\frac{1}{2})(-\frac{1}{2})(-\frac{1}{2})=\frac{1}{16}$

$(b) =(-\frac{1}{2})^{-4}=\frac{1}{(-\frac{1}{2})^{4}}=\frac{1}{(-\frac{1}{2})(-\frac{1}{2})(-\frac{1}{2})(-\frac{1}{2})}=\frac{1}{\frac{1}{16}}=16$

is the greater amount.

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Question

Which quantity is greater?

$(a)\ (3)^{-2}$

$(b)\ (\frac{1}{3})^{2}$

Answer

$(a)= (3)^{-2}=\frac{1}{3^{2}}=\frac{1}{9}$

$(b)=(\frac{1}{3})^{2}=(\frac{1}{3})(\frac{1}{3})=\frac{1}{9}$

and are equal.

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Question

Which quantity is greater?

$(a)\ -4(x^{2}+2x-10)$

$(b)\ -4x^2-8x-40$

Answer

$(a)= -4(x^{2}+2x-10)$

Use the distributive property to expand.

To compare the expressions, we can remove common factors. and are common in both expressions, so they can be removed:

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Question

Which quantity is greater?

$(a)\ -3(x^3+1)$

$(b)\ 3(-x^3+1)$

Answer

Use the distributive property to evaluate and .

Since , we get:

is the greater quantity.

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Question

Which quantity is greater?

$(a)\ -2(x^4+x^2-7)$

$(b)\ 2(-x^4-x^2+6)$

Answer

Use the distributive property to evaluate each quantity.

$(a) =-2(x^{4}+x^2-7)$

and are common in both expressions, so we can write:

is the greater quantity

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Question

Which quantity is greater?

$(a)\ (9)^{-2}$

$(b)\ (-3)^{-4}$

Answer

$(a)\ (9)^{-2}=\frac{1}{9^2}=\frac{1}{81}$

$(b)\ (-3)^{-4}=\frac{1}{(-3)^4}=\frac{1}{(-3)(-3)(-3)(-3)}=\frac{1}{81}$

and are equal.

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Question

Which of the following is equivalent to ?

Answer

We demonstrate that all four of the expressions are equivalent to .

By the associative property of multiplication,

$4 (5y) = (4 \cdot 5) \cdot y = 20y$

and

$5 (4y) = (5\cdot 4) \cdot y = 20y$

By combining like terms, and applying the distribution property:

$5(3y + y) = 5(3y + 1 y) = 5[ (3 + 1 ) y] = 5 \left ( 4 y \right )= 20y$

$4 (3y + 2y) = 4\left [ (3 + 2) y \right ] = 4\left ( 5y\right ) = 20y$

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Question

Which of the following is equivalent to $6t \cdot 7t$ ?

Answer

Use the commutative and associative properties to reorder the expression and group like-terms together.

$6t \cdot 7t = 6\cdot 7\cdot t \cdot t = 42 \cdot t^{2}=42t^2$

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Question

$\frac{\begin{array}[b]{r}10\ \times\ 1\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$1\times0=0$

$1\times1=1$

$\frac{\begin{array}[b]{r}10\ \times\ 1\end{array}}{ \ \ \space10}$

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Question

$\frac{\begin{array}[b]{r}20\ \times\ 2\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$2\times0=0$

$2\times2=4$

$\frac{\begin{array}[b]{r}20\ \times\ 2\end{array}}{ \ \ \space40}$

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Question

$\frac{\begin{array}[b]{r}30\ \times\ 3\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$3\times0=0$

$3\times3=9$

$\frac{\begin{array}[b]{r}30\ \times\ 3\end{array}}{ \ \ \space90}$

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Question

$\frac{\begin{array}[b]{r}40\ \times\ 4\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$4\times0=0$

$4\times4=16$

$\frac{\begin{array}[b]{r}40\ \times\ 4\end{array}}{ \ \space160}$

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Question

$\frac{\begin{array}[b]{r}50\ \times\ 5\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$5\times0=0$

$5\times5=25$

$\frac{\begin{array}[b]{r}50\ \times\ 5\end{array}}{ \ \space250}$

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Question

$\frac{\begin{array}[b]{r}60\ \times\ 6\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$6\times0=0$

$6\times6=36$

$\frac{\begin{array}[b]{r}60\ \times\ 6\end{array}}{ \ \space360}$

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Question

Solve the following:

$\frac{\begin{array}[b]{r}70\ \times\ 7\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$7\times0=0$

$7\times7=49$

$\frac{\begin{array}[b]{r}70\ \times\ 7\end{array}}{ \ \space490}$

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Question

$\frac{\begin{array}[b]{r}80\ \times\ 8\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$8\times0=0$

$8\times8=64$

$\frac{\begin{array}[b]{r}80\ \times\ 8\end{array}}{ \ \space640}$

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Question

$\frac{\begin{array}[b]{r}90\ \times\ 9\end{array}}{ \ \ \ \space}$

Answer

To solve this problem, we multiply by the ones position and the tens position.

$9\times0=0$

$9\times9=81$

$\frac{\begin{array}[b]{r}90\ \times\ 9\end{array}}{ \ \space810}$

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