The solutions to GR3768

 Posts: 3
 Joined: Tue Jun 25, 2024 6:42 pm
The solutions to GR3768
The newly released practice math Subject Test is very difficult.Is there any solution manual to this available??
Re: The solutions to GR3768
Try here. https://www.mathsub.com/solutions/
I also found this answer on the internet, not sure if its relevant:
We have realvalued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:
[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]
where (f) and (g) are realvalued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).
The correct answer is:
[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u}  \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]
I also found this answer on the internet, not sure if its relevant:
We have realvalued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:
[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]
where (f) and (g) are realvalued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).
The correct answer is:
[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u}  \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]

 Posts: 8
 Joined: Thu Mar 23, 2023 1:48 pm
Re: The solutions to GR3768
There is no solution manual as far as I know. However, if there are specific questions you're having trouble with, maybe you can post them here and people can explain their solutions?

 Posts: 3
 Joined: Tue Jun 25, 2024 6:42 pm
Re: The solutions to GR3768
Actually he has all previous GRE math solutions, except for the newly released ones .Also he is not responding to emails..This newly released sample test seems like it is a more realistic review of undergrad math , although I am not sure the actual test I will be taking will resemble anything like this.copilot wrote: ↑Fri Jun 28, 2024 11:01 amTry here. https://www.mathsub.com/solutions/
I also found this answer on the internet, not sure if its relevant:
We have realvalued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:
[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]
where (f) and (g) are realvalued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).
The correct answer is:
[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u}  \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]

 Posts: 8
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
I'm sure the test will be similar to 3768.
If you study all the previous exams there are some repeat/similar questions. For example there's always a question about dim(u+v) = dim(u) + dim(v)  dim(u intersect v)
there's usually a problem about isomorphic graphs
Etc
If you study all the previous exams there are some repeat/similar questions. For example there's always a question about dim(u+v) = dim(u) + dim(v)  dim(u intersect v)
there's usually a problem about isomorphic graphs
Etc

 Posts: 8
 Joined: Fri Aug 09, 2024 1:01 am
Re: The solutions to GR3768
For number 21, is this valid reasoning?
Add the two congruences together to get $4x + 9y \equiv 6 \mod 13$. Add this new congruence to the second one to get $5x + 16y \equiv 7 \mod 13$. Since $16y \equiv 3y$ we have that $5x + 3y \equiv 7 \mod 13$.
p.s. how to get LaTeX to render here????
Add the two congruences together to get $4x + 9y \equiv 6 \mod 13$. Add this new congruence to the second one to get $5x + 16y \equiv 7 \mod 13$. Since $16y \equiv 3y$ we have that $5x + 3y \equiv 7 \mod 13$.
p.s. how to get LaTeX to render here????