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ioda
2 years ago
5

When functions are defined by more than one​ equation, they are called?

Mathematics
1 answer:
USPshnik [31]2 years ago
3 0
When functions are defined by more than one equation, they are called piecewise- defined functions.

Hope that helped
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Brainliest if correct.
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Burt is thinking about a career in publishing but does not like writing. What
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A local company manufactures netbook computers. Their profit function is given by this equation: y equal minus x to the power of
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The equation described above can also be written as,                    y = -x² + 100x + 4000To get the number of notebooks that will give them the maximum profit, we derive the equation and equate to zero.                  dy/dx = -2x + 100 = 0The value of x from the equation is 50. Then, we substitute 50 to the original equation to get the profit.                    y = -(50^2) + 100(50) + 4000 = 6500Thus, the maximum profit that the company makes is $6,500/day. 
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3 years ago
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Find the second partial derivatives of the following functions
artcher [175]

(a) <em>z</em> = 3<em>x</em> ² - 4<em>xy</em> + 15<em>y</em> ²

has first-order partial derivatives

∂<em>z</em>/∂<em>x</em> = 6<em>x</em> - 4<em>y</em>

∂<em>z</em>/∂<em>y</em> = -4<em>x</em> + 30<em>y</em>

and thus second-order partial derivatives

∂²<em>z</em>/∂<em>x</em> ² = 6

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = -4

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = -4

∂²<em>z</em>/∂<em>y</em> ² = 30

where ∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = ∂/∂<em>x</em> [∂<em>z</em>/∂<em>y</em>] and ∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = ∂/∂<em>y</em> [∂<em>z</em>/∂<em>x</em>].

(b) <em>z</em> = 4<em>x</em> <em>eʸ</em>

∂<em>z</em>/∂<em>x</em> = 4<em>eʸ</em>

∂<em>z</em>/∂<em>y</em> = 4<em>x</em> <em>eʸ</em>

∂²<em>z</em>/∂<em>x</em> ² = 0

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = 4<em>eʸ</em>

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = 4<em>eʸ</em>

∂²<em>z</em>/∂<em>y</em> ² = 4<em>x</em> <em>eʸ</em>

<em />

(c) <em>z</em> = 6<em>x</em> ln(<em>y</em>)

∂<em>z</em>/∂<em>x</em> = 6 ln(<em>y</em>)

∂<em>z</em>/∂<em>y</em> = 6<em>x</em>/<em>y</em>

∂²<em>z</em>/∂<em>x</em> ² = 0

∂²<em>z</em>/∂<em>x</em>∂<em>y</em> = 6/<em>y</em>

∂²<em>z</em>/∂<em>y</em>∂<em>x</em> = 6/<em>y</em>

∂²<em>z</em>/∂<em>y</em> ² = -6<em>x</em>/<em>y</em> ²

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2 years ago
Help me pls imma fail
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Answer:

-

Step-by-step explanation:

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