Answer:
<B = 47°
<C = 28°
b = AC = 28.0
Step-by-step explanation:
Given:
∆ABC
AB = c = 18
BC = a = 37
<A = 105°
Required:
Length of AC = b
measure of angle B and angle C
SOLUTION:
==>Use the sine rule, sin A/a = sinC/c to find the angle of C:
SinA = sin(105) = 0.9659
a = 37
sinC = ?
c = 18
0.9659/37 = sinC/18
Cross multiply
0.9659*18 = 37*sinC
17.3862 = 37*sinC
Divide both sides by 37
17.3862/37 = sinC
0.4699 = sinC
sinC = 0.4699
C = Sin-¹(0.4699)
C = 28.0° (nearest tenth)
==>Find angle B using sum of angles in a triangle:
Angle B = 180 - (105+28)
Angle B = 180 - 133
Angle B = 47°
==>Find length of b using sine rule, b/sinB = c/sinC:
SinC = sin(28) = 0.4695
SinB = sin(47) = 0.7314
c = 18
b = ?
b/0.7314 = 18/0.4695
Cross multiply
b*0.4695 = 18*0.7314
b*0.4695 = 13.1652
Divide both sides by 0.4695
b = 13.1652/0.4695
b = 28.0 (nearest tenth)
Bandi jogged at a slower rate. Bandi’s rate: 12 minutes a mile Sharna’s rate: 12.8 minute miles
Answer:
87.5 square inches
Step-by-step explanation:
The belt knuckle as shown in the diagram above consist of a 4 rectangles (2 are the of the same dimension, the other two are of the same direction differently) and a rectangle.
Area of the belt knuckle = area of the 4 triangles + area of the rectangle
✔️Area of the 2 rectangles with the following dimensions:
base (b) = 9 inches
height (h) = 2.5 inches
Area of the two triangles = 2(½*b*h)
= 2(½*9*2.5)
= 22.5 inches²
✔️Area of the 2 rectangles with the following dimensions:
base (b) = 4 inches
height (h) = 5 inches
Area of the two triangles = 2(½*b*h)
= 2(½*4*5)
= 20 inches²
✔️Area of the rectangle = l*w
l = 9 inches
w = 5 inches
Area = 9*5 = 45 inches²
✔️Area of the belt knuckle = 22.5 + 20 + 45 = 87.5 square inches
Answer:
Here we will use the relation:
Density = mass/volume.
We know that:
Mass = 1485g
Volume = 750L
Density = 1485g/750L = 1.98 g/L
We do not have the options in order to select the correct one, but knowing this density, we can suppose the possible gases.
Two "common" gases with densities similar to this one are:
Carbon Dioxide, with d = 1.977 g/L
Dinitrogen Monoxide, with d = 1.977 g/L
In both cases, if we round up we will get the same density that we calculated at the beginning, so either of these can be the correct option.
A Beards, Beards where a fashion choice back then, just like how wemon had black teeth and white painted faces!
