Let the angles be x
We are taking both angles as x because they are equal!




➪ <em>T</em><em>h</em><em>u</em><em>s</em><em>,</em><em> </em><em>T</em><em>h</em><em>e</em><em> </em><em>m</em><em>e</em><em>a</em><em>s</em><em>u</em><em>r</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>b</em><em>o</em><em>t</em><em>h</em><em> </em><em>a</em><em>n</em><em>g</em><em>l</em><em>e</em><em>s</em><em> </em><em>i</em><em>s</em><em> </em><em>9</em><em>0</em><em>°</em><em>.</em><em>.</em><em>.</em><em>~</em>
Answer:
a. 215.6 in^3
b. 1.51 lb
Step-by-step explanation:
The area of each hand grip hole is that of a circle of radius 0.6 in together with a rectangle 2 in long and 1.2 in wide. So, that area is ...
π·(0.6 in)^2 + (2 in)(1.2 in) = (0.36π +2.4) in^2
The area of the kickboard before the hand grip holes are put in is that of a semicircle of radius 5.5 in together with a rectangle 12 in long and 11 in wide. So, that area is ...
(1/2)·π·(5.5 in)^2 + (12 in)(11 in) = (15.125π +132) in^2
Taking the hand grip holes out, the top area of the board is ...
((15.125π +132) -2(0.36π +2.4)) in^2
= (14.405π + 127.2) in^2
___
a. The volume is the product of the area and the thickness, so is ...
((14.405π +127.2) in^2)·(1.25 in) ≈ 215.568 in^3
__
b. The weight of the kickboard is the product of its volume and its density:
(215.568 in^3)(0.007 lb/in^3) ≈ 1.509 lb
Answer:
42 slices per hour
Step-by-step explanation:
IQR= Q3-Q1
Airport 1
Q3 is 8
Q1 is 2
8-2= 6
Airport 1 IQR is 6
Airport 2
Q3 is 8
Q1 is 3
8-3= 5
Airport 2 IQR Is 5
Answer:
Step-by-step explanation:
what?