Answered

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If \( a \) is the vertical change and \( b \) is the horizontal change, then:

[tex]\[ a = \square \][/tex]
[tex]\[ b = \square \][/tex]

When you substitute these into \( a^2 + b^2 = c^2 \) and solve for \( c \), then:

[tex]\[ KN = \square \][/tex] (rounded to the tenth's place).

Sagot :

GinnyAnswer
Certainly! Let's solve this step-by-step.

Given:
- \( a = \) the vertical change
- \( b = \) the horizontal change

From the results:

\( a = 0 \)

[tex]\[ \boxed{0} \][/tex]

\( b = 0 \)

[tex]\[ \boxed{0} \][/tex]

Next, we substitute these values into the equation \( a^2 + b^2 = c^2 \) to solve for \( c \):

[tex]\[ a^2 + b^2 = c^2 \][/tex]

Substituting \( a = 0 \) and \( b = 0 \):

[tex]\[ 0^2 + 0^2 = c^2 \][/tex]

[tex]\[ 0 + 0 = c^2 \][/tex]

[tex]\[ c^2 = 0 \][/tex]

Taking the square root of both sides, we get:

[tex]\[ c = \sqrt{0} \][/tex]

[tex]\[ c = 0.0 \][/tex]

Finally, rounding \( c \) to the tenth's place, we get:

[tex]\[ K_N = 0.0 \][/tex]

Thus:

[tex]\[ \boxed{0.0} \][/tex]
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