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Home » The function \( f(x) = \frac2x - 1x + 4 \) has a horizontal asymptote at what value? - jntua results

The function \( f(x) = \frac2x - 1x + 4 \) has a horizontal asymptote at what value? - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Let the integers be \(x\), \(x+2\), \(x+4\).

📌 x + (x+2) + (x+4) = 102 \quad \Rightarrow \quad 3x + 6 = 102 \quad \Rightarrow \quad 3x = 96 \quad \Rightarrow \quad x = 32

📌 Largest integer: \(32 + 4 = 36\)

📌 #### 36Question: A startup is developing a medical device that uses vector-based sensor arrays to detect physiological signals. Suppose two sensor vectors are given as $\vec{v} = \begin{pmatrix} 3 \\ -2 \\ x \end{pmatrix}$ and $\vec{w} = \begin{pmatrix} x \\ 4 \\ -1 \end{pmatrix}$. Find the value of $x$ such that the vectors are orthogonal.

📌 Solution: For two vectors to be orthogonal, their dot product must be zero. Compute the dot product:

📌 \vec{v} \cdot \vec{w} = 3x + (-2)(4) + x(-1) = 3x - 8 - x = 2x - 8

📌 Set the dot product equal to zero:

📌 2x - 8 = 0 \quad \Rightarrow \quad 2x = 8 \quad \Rightarrow \quad x = 4

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So, \( a = 1 \). Thus:
\frac{x + 2}{x - 2} = 1 \quad \Rightarrow \quad x + 2 = x - 2,
which simplifies to \( 2 = -2 \), a contradiction. This implies no sol...
But solving directly from the original:
Multiply both sides by \( (x - 2)(x + 2) = x^2 - 4 \):
2x^2 + 8 = 2x^2 - 8.
Subtract \( 2x^2 \):
a contradiction. Hence, no real solution exists. The equation is undef...
Compute the sum of the roots of the equation \( u\sqrt{u} - 5u + 6\sqr...
Let \( v = \sqrt{u} \), so \( u = v^2 \). Substitute:

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