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pentagon [3]
3 years ago
5

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Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0

\dfrac{\dfrac{x}{4}+\dfrac{1}{8}}{\dfrac{x^2}{4}}=\left(\dfrac{2\cdot x}{2\cdot4}+\dfrac{1}{8}\right):\dfrac{x^2}{4}=\left(\dfrac{2x}{8}+\dfrac{1}{8}\right)\cdot\dfrac{4}{x^2}\\\\=\dfrac{2x+1}{8}\cdot\dfrac{4}{x^2}=\dfrac{2x+1}{2}\cdot\dfrac{1}{x^2}=\dfrac{2x+1}{2x^2}\to A.

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kodGreya [7K]
<h3>Answer:  orthogonal </h3>

===========================================================

Explanation:

For any two vectors defined as follows

u = <a,b>

v = <c,d>

the dot product is computed by

u dot v = a*c + b*d

If the dot product of the vectors is 0, then the vectors are orthogonal. Meaning they are perpendicular to one another.

-------------------

Let's find the dot product of these two given vectors

u = < 2, -4 >

v = < 6, 3 >

u dot v = 2*6 + (-4)*3

u dot v = 12 - 12

u dot v = 0

Therefore, these two vectors form a right angle and are <u>orthogonal</u>

-------------------

Extra info:

If we can show that u = <a, b> and v = <ka, kb> for some real number k, then we have shown that vectors u and v are parallel.

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Step-by-step explanation:

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Answer:

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