Use elimination to solve the following system of equations: 17x - 2y = 25,17x + 3y = 5
1 answer:
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Step-by-step explanation:
so we know X= Numbers of Large boxes and Y= Numbers of Small boxes
And we know the large boxes weigh <em>7</em><em>5</em><em><u> </u></em><em><u>pounds</u></em> and the small boxes weigh <em>4</em><em>0</em><em> </em><em><u>pounds</u></em>
So I would have to say the the same except you have to flip the inequality sign like this:
75x + 40y
200
And if that doesnt somehow work and the question is wording it wrong then
My guess for why its wrong us because its not in slope intercept form Although you still can solve for either varible ( x or y) using standard form also.
So to get from standard form to Slope intercept form (y=mx+b) these are the steps:
Ax+by=C
75x + 40y ≤ 200
Turn it into a linear equation.
75x+ 40y =200
In order to go from one form to another, all you have to do is change the order of the given numbers. First you want to move the Ax to the opposite side of the equation, by either adding or subtracting it. At this point your equation will be set up By = -Ax + C. Then you want to divide the B from the By and the rest of the equation. Therefore you will have y = - Ax/B + C/B. This is the same thing as the slope-intercept form, just a few of the letters are different.
40y=-75x+200 first subtract 75x
y=−1.875×+5 then dived every varible (everything) by 40. and you have your Linear eqaution.
And your second question would be <em><u>A</u></em><em><u>.</u></em><em><u> </u></em><em><u>>The number of boxes must be a whole number.</u></em><em><u> </u></em>
Because you cannot split boxes in half or in any quarter in a real life scenario.
The given function is
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)