Let the weight of triangle be x units. You are given that the weight of square is 1 unit.
1) On the left side of the balanced beam you can see 3 triangles and 5 squares. The weight on left side is 3·x+5·1=3x+5 units.
2) On the right side of the balanced beam you can see 2 triangles and 7 squares. The weight on right side is 2·x+7·1=2x+7 units.
3) If the whole system is balanced, then the weights on left and right sides are equal:
3x+5=2x+7.
Solve this equation:
3x-2x=7-5,
x=2 units.
Answer: option A
<span>3
--- </span>-2 = 8<span>
y
</span>3
--- = 8+2
y
3
--- = 10
y
3
3
y = ------
10
Answer:
C) 45 ft²
Step-by-step explanation:
<u>Formulae</u>
Area of a square = x² (where x is the side length)
Area of a triangle = 1/2 × base × height
The surface area of a square pyramid comprises:
- square base
- 4 congruent triangles
⇒ area of square base = 5² = 25 ft²
⇒ area of triangular face = 1/2 × 5 × 2 = 5 ft²
⇒ Total surface area = area of square base + 4 triangular areas
= 25 + 4(5)
= 25 + 20
= 45 ft²
X=-3 Y=8 so (-3,8) is your final answer
Answer:
I believe C
Step-by-step explanation:
hope this helps
