Answer:
P = 70 mi A = 210 mi^2
Step-by-step explanation:
Sides:
a = 29 m
b = 20 m
c = 21 m
Angles:
A = 90 °
B = 43.6028 °
C = 46.3972 °
Other:
P = 70 m
s = 35 m
K = 210 mi^2
r = 6 m
R = 14.5 m
Agenda:
A = angle A
B = angle B
C = angle C
a = side a
b = side b
c = side c
P = perimeter
s = semi-perimeter
K = area
r = radius of inscribed circle
R = radius of circumscribed circle
SSS is Side, Side, Side
Heron’s formula says that if a triangle ABC has sides of lengths a, b, and c opposite the respective angles, and you let the semiperimeter, s, represent half of the triangle’s perimeter, then the area of the triangle is
Im 100% sure that the answer is:
Given:$75
25 hours
Answer:
(3+p)
Step-by-step explanation:
solution
first expression=9-P^2
=3^2 - P^2
=(3+P) (3-p)
second expression=p^2 + 3p
= p(3+p)
Therefore lCM= (3+p)
Answer:



Step-by-step explanation:
Given

See attachment
Solving (a): 
To solve for
, we make use of:

The relationship between both angles is that they are complementary angles
Make
the subject

Substitute
for 


Solving (b): 
To solve for
, we make use of:
The relationship between both angles is that they are complementary angles

Solving (c): 
To solve for
, we make use of:

The relationship between both angles is that they are alternate exterior angles.
So:

Answer:
it is correct they weigh the same
Step-by-step explanation: