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densk [106]
3 years ago
15

How do you explain -2≤2x-4<4

Mathematics
2 answers:
maxonik [38]3 years ago
3 0
-2\leq2x-4 < 4\ \ \ \ \ |add\ 4\\\\2\leq2x < 8\ \ \ \ |divide\ by\ 2\\\\1\leq x < 4

Answer:\boxed{1\leq x < 4\Rightarrow x\in(-\infty;\ 1]\ \cup\ (4;\ \ibfty)}
MrRissso [65]3 years ago
3 0
-2\ \textless \ 2x-4\ \textless \ 4 

&#10;-2+4\ \textless \ 2x\ \textless \ 4+4&#10; &#10;  | Add \ 4 \ to \ the \ whole \ equation| 

&#10;2\ \textless \ 2x\ \textless \ 8&#10; &#10; 

1\ \textless \ x\ \textless \  \dfrac{8}{2}   |Divide \ the \ whole \ equation \ by \ 2|

&#10;1\ \textless \ x\ \textless \ 4&#10; &#10;

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An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i
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The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

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The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

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The total time, 't', is given as follows;

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

The minimum value of 't' is found by differentiating with an online tool, as follows;

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

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x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

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