MHB Using linear systems to solve problems (2)

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Provincial Express charges a fixed fee of $4 plus a variable cost of $1.50 per kilogram for overnight delivery, while The Package People has a fixed fee of $5 and a variable cost of $1 per kilogram. To determine when Provincial Express becomes the cheaper option, one can set up linear equations based on the costs. The equation for Provincial Express is represented as $y = 1.5x + 4$, where "x" is the weight of the package. The discussion emphasizes the use of linear systems to analyze and compare these delivery costs effectively.
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Provincial Express charges \$4, plus \$1.50/kg, to deliver a package overnight. The Package people charges \$5, plus \$1/kg. When is Provincial Express less expensive to use?
 
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Exactly the same as the rental car. Name stuff and demonstrate what you can do.
 
What do you know about linear equations? This one will be of the form $y=mx+c$ where "c" is fixed cost and "m" being variable cost (in this case per kilo) and x is mass.

Provincial Express is $4+ 1.50x$ - 4 is the fixed cost and 1.5 the variable cost.

Can you set up an equation for Tehe?
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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