The heat loss h (in watts) through a single-pane glass window varies jointly with the window's area A (in square meters) and the difference between the inside and outside temperatures d (in kelvins). Write an equation relating h, A, d, and a constant k. If a single-pane of glass with an area of 1.00 square meter and a temperature difference of 1.00 kelvin has a heat loss of 5.70 watts, then what is the heat loss through a single-pane window with an area of 2.50 square meters and a temperature difference of 20 kelvins?

285 watts

Step-by-step explanation:

The heat loss h (in watts) through a single-pane glass window varies jointly with the window's area A (in square meters) and the difference between the inside and outside temperatures d (in kelvins). The equation relating h, A, d, and a constant k would be

h = kAd

If a single-pane of glass with an area of 1.00 square meter and a temperature difference of 1.00 kelvin has a heat loss of 5.70 watts, it means that

5.7 = k × 1 ×1

k = 5.7/1 = 5.7

The equation becomes

h = 5.7Ad

Therefore, the heat loss through a single-pane window with an area of 2.50 square meters and a temperature difference of 20 kelvins would be

h = 5.7 × 2.5 × 20

h = 285 watts

1) c 2143.6 cm

2) c 113.0 m

3) a 33.5 ft

4)   d 2 cm

5) c 20 mm

6) 180 in

7) d 120 ft

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)