QuestionFind the value of $(x-2y)(x+2y)$ for $x=4$ and $y=\frac{1}{2}$ . Choices: 8, 9, 14, 15.

Studdy Solution

STEP 1

Assumptions1. We are given an expression $(x- y)(x+ y)$ . We are given the values of $x$ and $y$ , which are $4$ and $1 /$ respectively.

STEP 2

We need to substitute the given values of $x$ and $y$ into the expression. $(x-2 y)(x+2 y) = (4 -2 \cdot \frac{1}{2})(4 +2 \cdot \frac{1}{2})$

STEP 3

Now, we simplify the expression inside the brackets.
$= ( -1)( +1)$

STEP 4

Now, we calculate the values inside the brackets.
$= (3)()$

STEP 5

Finally, we multiply the values to get the result.
$=15$ The value of the expression $(x-2 y)(x+2 y)$ when $x=4$ and $y=1 /2$ is $15$ .

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QuestionFind the value of $(x-2y)(x+2y)$ for $x=4$ and $y=\frac{1}{2}$ . Choices: 8, 9, 14, 15.

Studdy Solution

STEP 1

Assumptions1. We are given an expression $(x- y)(x+ y)$ . We are given the values of $x$ and $y$ , which are $4$ and $1 /$ respectively.

STEP 2

We need to substitute the given values of $x$ and $y$ into the expression. $(x-2 y)(x+2 y) = (4 -2 \cdot \frac{1}{2})(4 +2 \cdot \frac{1}{2})$

STEP 3

Now, we simplify the expression inside the brackets.
$= ( -1)( +1)$

STEP 4

Now, we calculate the values inside the brackets.
$= (3)()$

STEP 5

Finally, we multiply the values to get the result.
$=15$ The value of the expression $(x-2 y)(x+2 y)$ when $x=4$ and $y=1 /2$ is $15$ .

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QuestionFind the value of (x−2y)(x+2y)(x-2y)(x+2y)(x−2y)(x+2y) for x=4x=4x=4 and y=12y=\frac{1}{2}y=21​. Choices: 8, 9, 14, 15.

Studdy Solution

STEP 1

Assumptions1. We are given an expression (x−y)(x+y)(x- y)(x+ y)(x−y)(x+y). We are given the values of xxx and yyy, which are 444 and 1/1 /1/ respectively.

STEP 2

We need to substitute the given values of xxx and yyy into the expression.(x−2y)(x+2y)=(4−2⋅12)(4+2⋅12) (x-2 y)(x+2 y) = (4 -2 \cdot \frac{1}{2})(4 +2 \cdot \frac{1}{2}) (x−2y)(x+2y)=(4−2⋅21​)(4+2⋅21​)

STEP 3

Now, we simplify the expression inside the brackets.=(−1)(+1) = ( -1)( +1) =(−1)(+1)

STEP 4

Now, we calculate the values inside the brackets.=(3)() = (3)() =(3)()

STEP 5

Finally, we multiply the values to get the result.=15 =15 =15The value of the expression (x−2y)(x+2y)(x-2 y)(x+2 y)(x−2y)(x+2y) when x=4x=4x=4 and y=1/2y=1 /2y=1/2 is 151515.

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QuestionFind the value of $(x-2y)(x+2y)$ for $x=4$ and $y=\frac{1}{2}$ . Choices: 8, 9, 14, 15.

Assumptions1. We are given an expression $(x- y)(x+ y)$ . We are given the values of $x$ and $y$ , which are $4$ and $1 /$ respectively.

We need to substitute the given values of $x$ and $y$ into the expression. $(x-2 y)(x+2 y) = (4 -2 \cdot \frac{1}{2})(4 +2 \cdot \frac{1}{2})$

Now, we simplify the expression inside the brackets.
$= ( -1)( +1)$

Now, we calculate the values inside the brackets.
$= (3)()$

Finally, we multiply the values to get the result.
$=15$ The value of the expression $(x-2 y)(x+2 y)$ when $x=4$ and $y=1 /2$ is $15$ .