Answer:
![e=\frac{1}{4} fg](https://tex.z-dn.net/?f=e%3D%5Cfrac%7B1%7D%7B4%7D%20fg)
Step-by-step explanation:
Joint variation problem solve using the equation y = kxz.
e ∝ fg
e=kfg
now substitute the values
![4=k*2*8\\4=16k\\k=\frac{4}{16} \\k=\frac{1}{4}](https://tex.z-dn.net/?f=4%3Dk%2A2%2A8%5C%5C4%3D16k%5C%5Ck%3D%5Cfrac%7B4%7D%7B16%7D%20%5C%5Ck%3D%5Cfrac%7B1%7D%7B4%7D)
The relationship will be:
![e=kfg\\e=\frac{1}{4} fg](https://tex.z-dn.net/?f=e%3Dkfg%5C%5Ce%3D%5Cfrac%7B1%7D%7B4%7D%20fg)
Answer:
B-is correct
Step-by-step explanation:
y=a(x-h)^2+k
-k -k
y-k=a(x-h)^2
:(x-h)^2 :(x-h)^2
(y-k)/(x-h)^2 =a
Answer:
Step-by-step explanation:
Good at what?
Answer:
Step-by-step explanation:
Since it is a regular hexagon, the length of its sides are equal. And same as the distance across its flats.
So that;
NR ≅ OP (property of a regular polygon)
PM ≅ RM (half of the distance across flats of a polygon)
NM ≅ OM (half of the distance across flats of a polygon)
<NMR ≅ <PMO (vertically opposite angles)
<NRM ≅ <OPM (alternate angle property)
<RNM ≅ <POM (alternate angle property)
This therefore proves that: ΔNRM = ΔOPM