Solve for d d = square root of (5--2)^2+(-4-1)^2

$d = \sqrt{{(5 - - 2)}^{2} + {(- 4 - 1)}^{2}}$

Step 1

Replace all occurrences of $-$ $-$ with a single $+$ . Two consecutive minus signs have the same mathematical meaning as a single plus sign because $- 1 \cdot - 1 = 1$

$d = \sqrt{{(5 + 2)}^{2} + {(- 4 - 1)}^{2}}$

Step 2

Add $5$ and $2$ .

$d = \sqrt{7^{2} + {(- 4 - 1)}^{2}}$

Step 3

Raise $7$ to the power of $2$ .

$d = \sqrt{49 + {(- 4 - 1)}^{2}}$

Step 4

Subtract $1$ from $- 4$ .

$d = \sqrt{49 + {(- 5)}^{2}}$

Step 5

Raise $- 5$ to the power of $2$ .

$d = \sqrt{49 + 25}$

Step 6

Add $49$ and $25$ .

$d = \sqrt{74}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$d = \sqrt{74}$

Decimal Form:

$d = 8.60232526 \dots$

Solve for d d = square root of (5--2)^2+(-4-1)^2

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