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balu736 [363]
1 year ago
5

Y varies inversely with x, y = 4 and x = 3, what is y, when x = 2B) given: y = 2x - 1 (x-3)(x+1) what is the I. VAII. HA III. X-

INTERCEPT

Mathematics
1 answer:
Semmy [17]1 year ago
4 0

SOLUTION

(a)

\begin{gathered} y\alpha\frac{1}{x} \\  \\ y=\frac{k}{x} \\  \\ 4=\frac{k}{3} \\  \\ k=12 \end{gathered}\begin{gathered} y=\frac{12}{x} \\  \\ when\text{ x=2} \\  \\ y=\frac{12}{2} \\  \\ y=6 \end{gathered}

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Answer:R(-10, 5.6) => R'(10, -5.6)

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Step-by-step explanation:

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What is the minimum value of the objective function, C with given constraints? C=5x+3y {⎨x+3y≤9 {5x+2y≤20 {x≥1 {y≥2
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Answer:

The answer is below

Step-by-step explanation:

Plotting the following constraints using the online geogebra graphing tool:

x + 3y ≤ 9         (1)

5x + 2y ≤ 20    (2)

x≥1 and y≥2      (3)

From the graph plot, the solution to the constraint is A(1, 2), B(1, 2.67) and C(3, 2).

We need to minimize the objective function C = 5x + 3y. Therefore:

At point A(1, 2): C = 5(1) + 3(2) = 11

At point B(1, 2.67): C = 5(1) + 3(2.67) = 13

At point C(3, 2): C = 5(3) + 3(2) = 21

Therefore the minimum value of the objective function C = 5x + 3y is at point A(1, 2) which gives a minimum value of 11.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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