32t + 148 = 0148 = 32t4.625 = tt ≈ 4.63h = -16 • (4.625)^2 + 148 • 4.625 + 30 = 372.25
ANSWER: is D. 4.63 sec; 372.25 ft
2.) 2x - 4 = 0 and 2x - 1 = 0x = {4/2, ½}x = {2,½} ANSWER: is B 2, 1 over 2
        
                    
             
        
        
        
Answer:
 = 5.74W
= 5.74W
 =0.27 W
=0.27 W
Explanation:
d= 2.6cm =>0.026m (for both the ceramic and the carpet )
Thermal conductivity of wool ' '=
'=  =
 =  =  0.04J/(sm °C)
=  0.04J/(sm °C) 
 Thermal conducticity of carpet ' ' = 0.84 J/(sm °C)
' = 0.84 J/(sm °C) 
Area 'A'=  =
=  = 77.2cm²=> 77.2 x
= 77.2cm²=> 77.2 x  m²
m²
 =33.0°C
=33.0°C
 =10.0°C
=10.0°C
Average Power  is determined by dividing amount of energy'Q' by time taken for the transfer't':
 is determined by dividing amount of energy'Q' by time taken for the transfer't':
 = Q/Δt
 = Q/Δt
Due to conductivity, heat of flow rate will be P= dQ/dt
P= =
 = ![\frac{kA[T_{h}-T_{c} ]}{d}](https://tex.z-dn.net/?f=%5Cfrac%7BkA%5BT_%7Bh%7D-T_%7Bc%7D%20%5D%7D%7Bd%7D)
For CERAMIC:
 =
=![\frac{k_{c} A[T_{h}-T_{c} ]}{d}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7Bc%7D%20A%5BT_%7Bh%7D-T_%7Bc%7D%20%5D%7D%7Bd%7D) => [0.84 x 77.2 x
 => [0.84 x 77.2 x  (33-10) ]/0.026
(33-10) ]/0.026
 = 5.74W
= 5.74W
For WOOL CARPET:
 =
= ![\frac{k_{w} A[T_{h}-T_{c} ]}{d}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7Bw%7D%20A%5BT_%7Bh%7D-T_%7Bc%7D%20%5D%7D%7Bd%7D) => [0.04 x 77.2 x
=> [0.04 x 77.2 x  (33-10) ]/0.026
(33-10) ]/0.026
 =0.27 W
=0.27 W