3 years ago

How do I solve this equation?

Mathematics

2 answers:

denis-greek [22]3 years ago

7 0

Answer:C:7

Step-by-step explanation:

svlad2 [7]3 years ago

5 0

The answer is option three(photo for step by step)

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Identify the congruent angles in the figure.<br> Explain your reasoning.

kirill115 [55]

Answer:2 and 4 and 1 and three because there in the same corners

Step-by-step explanation: 2 and 4 are in open corners while 3 and 1 are in closed corners

6 0

3 years ago

Read 2 more answers

Consider the vector field f(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)kf(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)k.

natima [27]

$\dfrac{\partial f}{\partial x}=2z+3y\implies f(x,y,z)=2xz+3xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=3x+\dfrac{\partial g}{\partial y}=3z+3x$
$\dfrac{\partial g}{\partial y}=3z\implies g(y,z)=3yz+h(z)$
$\implies f(x,y,z)=3xz+3xy+3yz+h(z)$

$\dfrac{\partial f}{\partial z}=3x+3y+\dfrac{\mathrm dh}{\mathrm dz}=3y+2x$
$\dfrac{\mathrm dh}{\mathrm dz}=-x$

But we assume $h$ is a function of $z$ alone, so there is no solution for $h$ and hence no scalar function $f$ such that $\nabla f=\mathbf f$ .