A) The new dimensions would be 32cm by 12cm.
B) This is an enlargement.
To find the new dimensions, multiply the original dimensions by the scale factor. 8*4=32 and 3*4=12.
This is an enlargement because the copy is larger than the original.
Answer:
9:34 pm
Step-by-step explanation:
First:
8:17 am
Go back 12 hours
8:17 pm
Round to 8:20
(keep the 3) Minus 43-17=26(3 is in this)
Therefore, 9:34 pm
Sorry if this doesn't make sense, it's hard to explain haha
Answer:
General Solution is
and the particular solution is 
Step-by-step explanation:

This is a linear diffrential equation of type
..................(i)
here 

The solution of equation i is given by

we have ![e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}](https://tex.z-dn.net/?f=e%5E%7B%5Cint%20p%28x%29dx%7D%3De%5E%7B%5Cint%20%5Cfrac%7B-2%7D%7Bx%7Ddx%7D%5C%5C%5C%5Ce%5E%7B%5Cint%20%5Cfrac%7B-2%7D%7Bx%7Ddx%7D%3De%5E%7B-2ln%28x%29%7D%5C%5C%5C%5C%3De%5E%7Bln%28x%5E%7B-2%7D%29%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%20%7D%20%5C%5C%5C%5C%5Cbecause%20e%5E%7Bln%28f%28x%29%29%7D%3Df%28x%29%5D%5C%5C%5C%5CThus%5C%5C%5C%5Ce%5E%7B%5Cint%20p%28x%29dx%7D%3D%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%7D)
Thus the solution becomes


This is the general solution now to find the particular solution we put value of x=2 for which y=6
we have 
Thus solving for c we get c = -1/2
Thus particular solution becomes

Simple,
writing out the problem...

Simplify it...
-4x(6+5-1)
-4x(11-1)
-4x(10)
-40x
Making it look like...

Cancel out the common factors...
Thus making the answer:

.