1. Find the H.C.F. of 4x2y3 and 6xy2z.
Solution:
The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.
Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31
Therefore, the H.C.F. of 4 and 6 is 2
54 = 2(4x-3) + 2(x)
54 = 8x-6 + 2x
60 = 10x
6 = x
so two sides are 6 meters, and the other 2 sides are 21 meters
Answer:
The answer is C becuase 18÷3=6
Answer / Step-by-step explanation:
It should be noted that the question is incomplete due to the fact that the diagram has not been provided. However, the diagram has been complementing the question has been provided below.
To solve the question in the narrative, we recall the equation used in solving for displacement:
Thus, δₙₐ = Σ pL/AE
Where:
P is applied axial force.
E is the young's modulus of elasticity.
A is the area of cross-section.
L is length of the bar
Therefore, -8 (80) ÷ π/4 ( 0.85)² (18) (10³) + 2(150) ÷ π/4 (1.1)² (18) (10³) + 6(100) ÷ π/4 (0.45)² (18) (10³)
Solving further,
we have,
-8 (80) ÷ 0.7853( 0.85)² (18) (10³) + 2(150) ÷ 0.7853(1.1)² (18) (10³) + 6(100) ÷ 0.7853 (0.45)² (18) (10³)
= -640÷ 0.7853( 0.85)² (18) (10³) + 300 ÷ 0.7853(1.1)² (18) (10³) + 600 ÷ 0.7853 (0.45)² (18) (10³)
Solving further, we arrive at 0.111 in answer.
The positive sign indicates that end A moves away from end D.
Answer:
8978.9
Step-by-step explanation:
MULTIPLY THE THE GIVEN AND
8.89 × 1010 = 8978.9
THEN THE RESULT WOUD BE THE STANDARD NOTATION.