Help plz so close to being done

Mathematics

2 answers:

Natasha2012 [34]3 years ago

8 0

Answer:

20 is the answer I got

algol [13]3 years ago

4 0

The answer for the area is 20

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Consider f and c below. f(x, y) = (3 + 4xy2)i + 4x2yj, c is the arc of the hyperbola y = 1/x from (1, 1) to 3, 1 3 (a) find a fu

irga5000 [103]

We want to find a scalar function $f(x,y)$ such that $\nabla f(x,y)=\mathbf f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j$ .
So we need to have
$\dfrac{\partial f}{\partial x}=3+4xy^2$
Integrating both sides with respect to $x$ gives
$f(x,y)=3x+2x^2y^2+g(y)$
Differentiating with respect to $y$ gives
$\dfrac{\partial f}{\partial y}=4x^2y+\dfrac{\mathrm dg}{\mathrm dy}=4x^2y$ $\implies\dfrac{\mathrm dg}{\mathrm dy}=0$ $\implies g(y)=C$
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4 years ago

8. A computer application generates a sequence of musical notes using the function f(n) 6(16)", where n is the

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Applying an exponential property, it is found that the function that will generate the same note sequence as function f(n) is given by:

B. $H(n) = 6(2)^{4n}$

<h3>What is function for the note sequence?</h3>

The function for the node sequence is defined by:

$f(n) = 6(16)^n$

A function that will the same note sequence as function f(n) has the same initial value of 6. Additionally, applying an exponential property, we have that:

$H(n) = 6(2)^{4n} = 6(2^4)^n = 6(16)^n$