Answer: radius r = 2 inches
height h = 6 inches
Step-by-step explanation:
Given;
Volume V = 24πin3
Volume of a cylinder is given by
V = πr^2h
h = V/πr^2. ....1
Where, h = height and r = radius of cylinder
For the surface area of the cylinder with open top. we have,
S = 2πrh + πr^2
For the cost of materials used, let k represent the cost of materials used for the body of the cylinder.
Then, for the bottom will be 3k
Total cost will be represented by C, which gives
C = 2πrhk + 3πr^2k. .....2
Substituting eqn 1 to 2, we have;
C = 2πrVk/πr^2 + 3πr^2k
C = 2Vk/r + 3πr^2k
The material cost is minimum at dC/dr = 0
dC/dr = -2Vk/r^2 + 6πrk =0
6πrk = 2Vk/r^2
r^3 = 2V/6π
r = (2×24π/6π)^-3
r = (8)^-3
r = 2
Substituting r = 2 into eqn1
h = 24π/π(2^2)
h = 24/4 = 6
h = 6
Lines <em>a</em> and <em>b</em> are parallel, so lines <em>p</em>, <em>q</em>, and <em>t</em> are considered to be transversals. To solve this, you make use of the fact that alternate interior angles are equal, as are alternate exterior angles, as are corresponding angles. Of course any linear pair of angles is supplementary.
∠1 = 90° — corresponding angle to the right angle above it
∠2 = 68° — the sum of 22° and angles 1 and 2 is 180°
∠3 = 112° — supplementary to angle 2 (and the sum of 22° and 90°, opposite interior angles of the triangle)
∠4 = 112° — equal to angle 3
∠5 = 68° — equal to angle 2; supplementary to angle 4
∠6 = 56° — base angle of isosceles triangle with 68° at the apex; the complement of half that apex angle
∠7 = 124° — supplementary to the other base angle, which is equal to angle 6; also the sum of angles 5 and 6
∠8 = 124° — alternate interior angle with angle 7, hence its equal.
lxwxh = v 2 x 2 x 2 = 8 7x8=56 Answer: 7
Answer:
6.31 mi
Step-by-step explanation:
The diagram below explains the solution better.
From the diagram,
C = starting point of the race.
A = end of the first part of the race.
B = end of the race.
Using Cosine rule, we can find the straight-line distance between the starting point and the end of the race.
Cosine rule states that:
![a^2 = b^2 + c^2 - 2bc[cos(A)]](https://tex.z-dn.net/?f=a%5E2%20%3D%20b%5E2%20%2B%20c%5E2%20-%202bc%5Bcos%28A%29%5D)
where A = angle A = <A
Given that
b = 5.2 miles
c = 2.0 miles
<A = 115° (from the diagram)
Hence,
![a^2 = 5.2^2 + 2.0^2 - 2*5.2*2.0[cos(115)]\\\\a^2 = 27.04 + 4 - 20.8[cos(115)]\\\\a^2 = 31.04 + 8.79\\\\a^2 = 39.83\\\\a = \sqrt{39.83}\\ \\a = 6.31 mi](https://tex.z-dn.net/?f=a%5E2%20%3D%205.2%5E2%20%2B%202.0%5E2%20-%202%2A5.2%2A2.0%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2027.04%20%2B%204%20-%2020.8%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2031.04%20%2B%208.79%5C%5C%5C%5Ca%5E2%20%3D%2039.83%5C%5C%5C%5Ca%20%3D%20%5Csqrt%7B39.83%7D%5C%5C%20%5C%5Ca%20%3D%206.31%20mi)
The straight-line distance between the starting point and the end of the race is 6.31 mi
