A. 1 = 3/3 = 4/4 = 52/52 = x/x
B. 5 = 10/2 = 15/3 = 7.5/1.5 = 120/24 = 5a/a
All the ratios must equal each other
As you can see in the picture above, there are six faces of a rectangular prism; two are formed with dimensions width and height, two are formed by the dimensions length and width, and two are formed by the dimensions length and height. So, if you know the length, width, and height of the rectangular prism, then the formula for the surface area is
=(2⋅ℎ⋅ℎ)+(2⋅ℎ⋅ℎℎ)+(2⋅ℎ⋅ℎℎ)
Answer:
10 times greater
Step-by-step explanation:
Because value of 5 in 503 is 5*100
Then the value of 5 in 53 is 5*10
So then value of 5 in 503 is 10 times greater than the value of 5 in 53
I hope this helps
Answer:
(A) 5.47 km
(B) 1.38 km
Step-by-step explanation:
The relationship between Pressure (mmHg) and Height (km) is given by the equation below:
p =760 e⁻ 0.145h----------------------------------------------------- (1)
(a) The height of the aircraft can be calculated by making 'h' the subject of equation (1)
In (p/760) = -0.145 h
h = -In (p/760) /0.145
h= In (760/p)/ 0.145------------------------------------------------- (2)
For the aircraft, p =344 mmHg; substituting into (2)
h = In (760/344)/0.145
= 0.7926/0.145
= 5.4662 km
≈ 5.47 km
(b) p= 622 mmHg at the mountain top. Substituting into equation (2) we have:
h = In (760/622)/0.145
= 0.2003 / 0.145
= 1.3813 km
≈ 1.38 km
The length of the diagonal of the canvas is approximately 27 degrees.
The height of the rectangular canvas must reach 18 inches. It must form a 48 degrees angle with the diagonal at the top of the canvas.
<h3>Length of the diagonal Canvas</h3>
Therefore, the length of the diagonal can be found as follows:
Using trigonometric ratio,
- cos ∅ = adjacent / hypotenuse
where
∅ = 48°
adjacent side = Height of the rectangle = 18 inches
hypotenuse = Length of the diagonal
Therefore,
cos 48° = 18 / h
cross multiply
h = 18 / cos 48°
h = 18 / 0.66913060635
h = 26.9005778976
length of the diagonal ≈ 27 inches
learn more on rectangle here: brainly.com/question/26099609?referrer=searchResults
