Answer:
To determine the number of real number solutions of as system of equations in which one equation is linear and the other is quadratic
1) Given that there are two variables, x and y as an example, we make y the subject of the equation of the linear equation and substitute the the expression for y in x into the quadratic equation
We simplify and check the number of real roots with the quadratic formula, 
for quadratic equations the form 0 = a·x² - b·x + c
Where b² > 4·a·c there are two possible solutions and when b² = 4·a·c equation there is only one solution.
Step-by-step explanation:
Angle AQB is x = 90
Angle ASB is x = 90
Angle ALB is x = 90
Angle ATB is x = 90
Angle ARB is x = 90
<span>BWD is x < 90</span>
Answer:
All the expressions other than option E, is equivalent to the expression 18m - 12.
Step-by-step explanation:
A. 6m - 4 + 6m -4 + 6m - 4
or 6m+6m+6m -4 -4 -4
or 18m -12
B. 12m + 6 - 6m -6
or 12m - 6m + 6 - 6
or 6m
C.6(3m - 2)
or 18m - 12
D.3(6m - 4)
or 18m - 12
E. 24n - 4² + 8 -6m
This option can not satisfy the given expression as it contains another variable as n.
The vertical numbers are the distance and the horizontal numbers are the time in seconds.
Look at where the line is located at the number 20.
It si between the 2 and the 3 on the horizontal line, but you can see it is closer to the 2 than it is the 3, so the time would be 2.2s
Step 1:
<span>Calculate the effective thermal conductivity of the wall or ceiling:
</span>
K_eff = [ (13 ÷ 8)(0.12) + (16 - (13 ÷ 8)) × (0.04)] ÷ 16
K_eff =<span> [ 0.195 + 0.565] </span>÷<span> 16
</span>
K_eff = 0.76 ÷ 16
K_eff = 0.0475 W/ (m K)
Step 2:
Calculate <span>the interior ceiling area:
</span>Area of each of the interior side walls = <span>8.82 m x 8.64 m
= 76.2 m</span>²
Area of the interior ceiling = 8.64 m × <span>8.64 m
</span> = 74.6 m²
H = - k·A·(Δ - T) ÷ <span>(thickness)
</span>
H = - 0.0475 ÷ (379.45 × 20) ÷ 45/8
H = - ( - 0.95 × 379.45 ) ÷<span> 0.1429
</span>
H = <span>2.52 kW </span>
