Answer:
First Question:
A) -15+20x B) 36x+28
Second Question:
-10/5, 1/25, 1/2, 3/5, 5/8
.we can call segment "a" the radius, and if we do that, we calculate the area of the triangle as follows:
Look at the vertical line bisecting the triangle, made up of segments of length "a" and "h"
Now look at the smaller, right triangle towards the bottom left of the equilateral triangle, with sides "a", "h", and "S/2".
If we're taking "a" to be the radius (in our case 6 inches), then
h
a
=
sin
30
degrees, so
h
6
=
0.5
, so h = 3 inches. Therefore the height of the equilateral triangle is
6
+
3
=
9
inches.
The original equilateral triangle in the figure is bisected by the vertical line, making 2 right triangles of height 9 inches and a base of length S/2.
S
2
a
=
cos
30
, so
S
2
=
0.866
⋅
6
=
5.196
inches (rounding).
So, the original equilateral triangle in the figure is twice the area of this larger right triangle. A right triangle's area is
1
2
b
⋅
h
. We have TWO of them, so the total area is
2
⋅
1
2
b
⋅
h
=
5.196
⋅
9
...which works out to:
46.77 square inches (rounding)
GOOD LUCK
Answer:
b=197.6cm
Step-by-step explanation:
Given that a = 41 and B =78°
From the diagram of the triangle,
Tan θ = opp/adj
Where θ=78°
Tan78°= b/41
b = tan78×41
b = 4.70× 41
b = 197.5944cm
To the nearest tenth
b=197.6cm
The answers are f = 2 and p = -2