Answer:
the tent can hold 8.66 lb of water (or 240 in³ of it)
Step-by-step explanation:
assuming that the tent base is a rectangular area , and the vertical cross section is triangular , then the volume of the tent will be:
Volume of the tent = Area* height/2 = 60 in² * 8 in / 2 = 240 in³
assuming the density of water as 62.4 lb/ft³ then
mass of water = density * volume = 62.4 lb/ft³ * 240 in³ * (1 ft / 12 in)³ = 8.66 lb
then the tent can hold 8.66 lb of water (or 240 in³ of it)
I find that turning the fractions into decimals helps in this situation.
9/16 = 0.5625
8/15 = 0.5333
so 9/16 is greater then 8/15
(-1, 3)
this is because in a coordinate plane, using the coordinates you gave it is the missing vertice is (-1,3)
U(x) = f(x).(gx)
v(x) = f(x) / g(x)
Use chain rule to find u(x) and v(x).
u '(x) = f '(x) g(x) + f(x) g'(x)
v ' (x) = [f '(x) g(x) - f(x) g(x)] / [g(x)]^2
The functions given are piecewise.
You need to use the pieces that include the point x = 1.
You can calculate f '(x) and g '(x) at x =1, as the slopes of the lines that define each function.
And the slopes can be calculated graphycally as run / rise of each graph, around the given point.
f '(x) = slope of f (x); at x = 1, f '(1) = run / rise = 1/1 = 1
g '(x) = slope of g(x); at x = 1, g '(1) = run / rise = 1.5/ 1 = 1.5
You also need f (1) = 1 and g(1) = 2
Then:
u '(1) = f '(1) g(1) + f(1) g'(1) = 1*2 + 1*1.5 = 2 + 1.5 = 3.5
v ' (x) = [f '(1) g(1) - f(1) g(1)] / [g(1)]^2 = [1*2 - 1*1.5] / (2)^2 = [2-1.5]/4 =
= 0.5/4 = 0.125
Answers:
u '(1) = 3.5
v '(1) = 0.125
