Answer:
The graph representing the linear inequalities is attached below.
Explanation:
The inequalities given are :
y>x-2 and y<x+1
For tables for values of x and y and get coordinates to plot for both equation.
In the first equation;
y>x-2
y=x-2
y-x = -2
The table will be :
x y
-2 -4
-1 -3
0 -2
1 -1
2 0
The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)
Use a dotted line and shade the part right hand side of the line.
Do the same for the second inequality equation and plot then shade the part satisfying the inequality.
The graph attached shows results.
Answer: y = x * 1dollars - 30dollars
Explanation:
Giving that the delivery cost in dollar is potent for all x > = 50 pounds of wght
Y = (x - 50)*1 dollar + c ...equ 1
Y = delivery cost equation in dollars
x = weigt of baggage for delivery
c = 20dollars = down payment for the first 50 pound weight of baggage
Equ 1 becomes
Y = (x)dollars - 50 dollars + 20 dollars
Y = (x) dollars -30 dollars
Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W
Answer:
q = 1.73 W
Explanation:
given data
small end = 5 cm
large end = 10 cm
high = 15 cm
small end is held = 600 K
large end at = 300 K
thermal conductivity of asbestos = 0.173 W/mK
solution
first we will get here side of cross section that is express as
...............1
here x is distance from small end and S1 is side of square at small end
and S2 is side of square of large end and L is length
put here value and we get
S = 5 + 
S = 
m
and
now we get here Area of section at distance x is
area A = S² ...............2
area A = 
m²
and
now we take here small length dx and temperature difference is dt
so as per fourier law
heat conduction is express as
heat conduction q = 
...............3
put here value and we get
heat conduction q = 
it will be express as
now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K
solve it and we get
q (30) = (0.173) × (600 - 300)
q = 1.73 W
