If \(\left\{ \begin{matrix} ''⊕\;''~means~''-\;'', \\ ''⊗\;''~means~''\div\;'', \\ ''\text{ }\!\!\;Δ\;\!\!\text{ }''~means~''+\;'', \\ ''\;∇\;''~means~''\times\;'', \\ \end{matrix} \right.\) then, the value of the expression Δ 2 ⊕ 3 Δ ((4 ⊗ 2) ∇ 4) =
7
To accurately evaluate the given mathematical expression, we first need to understand the mapping of the custom symbols to their standard arithmetic operators. This problem tests our ability to interpret given rules and apply the correct order of operations (BODMAS/PEMDAS) in a step-by-step manner.
The question defines the following transformations for the custom symbols into standard mathematical operators:
| Symbol | Standard Operator | Meaning |
|---|---|---|
| ⊕ | - | Subtraction |
| ⊗ | ÷ | Division |
| Δ | + | Addition |
| ∇ | × | Multiplication |
The given expression is: \( \Delta 2 \oplus 3 \Delta ((4 \otimes 2) \nabla 4) \)
Let's evaluate this expression by replacing the symbols with their corresponding operators and following the order of operations.
The expression \( \Delta 2 \oplus 3 \Delta ((4 \otimes 2) \nabla 4) \) becomes:
\( 2 - 3 + ((4 \div 2) \times 4) \)
Calculate the term \( (4 \div 2) \):
\( 4 \div 2 = 2 \)
Now, substitute this value back into the expression:
\( 2 - 3 + (2 \times 4) \)
Calculate the term \( (2 \times 4) \):
\( 2 \times 4 = 8 \)
Substitute this value back into the expression:
\( 2 - 3 + 8 \)
First, calculate \( 2 - 3 \):
\( 2 - 3 = -1 \)
The expression simplifies to:
\( -1 + 8 \)
Calculate \( -1 + 8 \):
\( -1 + 8 = 7 \)
After performing all the symbol substitutions and arithmetic operations in the correct order, the value of the expression \( \Delta 2 \oplus 3 \Delta ((4 \otimes 2) \nabla 4) \) is 7.
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :