keeping in mind that in a parallelogram the diagonals bisect each other, namely cut each other into two equal halves. Check the picture below.
![\stackrel{GH}{3x-4}~~ = ~~\stackrel{HE}{5y+1}\implies 3x=5y+5\implies x=\cfrac{5y+5}{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{DH}{x+1}~~ = ~~\stackrel{HF}{3y}\implies \stackrel{\textit{substituting "x" in the equation}}{\cfrac{5y+5}{3}+1~~ = ~~3y}](https://tex.z-dn.net/?f=%5Cstackrel%7BGH%7D%7B3x-4%7D~~%20%3D%20~~%5Cstackrel%7BHE%7D%7B5y%2B1%7D%5Cimplies%203x%3D5y%2B5%5Cimplies%20x%3D%5Ccfrac%7B5y%2B5%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BDH%7D%7Bx%2B1%7D~~%20%3D%20~~%5Cstackrel%7BHF%7D%7B3y%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20%22x%22%20in%20the%20equation%7D%7D%7B%5Ccfrac%7B5y%2B5%7D%7B3%7D%2B1~~%20%3D%20~~3y%7D)

Answer: points for points
Step-by-step explanation:
Hi t sure what to do?
Answer: 8n
Step-by-step explanation: 5n + 3n
Answer:
a.) b⁷÷b⁴
= <u>b⁷</u>
b⁴
= b^7-4
= b³
b.) <u>x </u><u> </u><u>×</u><u> </u><u> </u><u>x⁵</u>
x² × x
= <u>x</u><u>^</u><u>1+</u><u>5</u>
x^2+1
= <u>x⁶</u>
x³
= x^6-3
= x³

Let

The curl is

where
denotes the partial derivative operator with respect to
. Recall that



and that for any two vectors
and
,
, and
.
The cross product reduces to

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

which means
is indeed conservative and we can find
.
Integrate both sides of

with respect to
and

Differentiate both sides with respect to
and




Now

and differentiating with respect to
gives




for some constant
. So
