JEE Main Important Physics formulas
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In the maths table, students learn how many times a given number can be written. We have been taught multiplication facts ever since primary school. These maths tables make it easier for us to quickly solve problems involving multiplication. The maths tables 1 through 20 must therefore be committed to memory for students to be able to complete challenging calculations with ease. Let's review the maths tables from 1 to 20 here and discuss how to memorise them. Mathematical proficiency with maths tables is a fundamental building block. The gate to multi-digit multiplication is opened by it. It establishes the groundwork for algebra, long division, and fractional simplifying. The area model of multiplication becomes increasingly important in higher grades. The key to becoming a successful maths learner is for the students to memorise these tables and get past the obstacles that have been shown to prevent success in life, education, and employment.
The number 113 is prime. Only the number itself and the number 1 can divide the number 113. A number needs to have precisely two factors in order to be categorised as a prime number. 113 is a prime number because it has exactly two factors, 1 and 113.
So, when a number n is added to itself ‘m’ times the result is equal to the number as “n ✖ m”. As if I wanted the number 113 to be multiplied by three, as in 113+113+113=339.
Therefore, adding the same natural number to the desired number of times yields the answer of 339, or I can simply multiply the same natural number by the appropriate number of times, which comes out to be 3 in this case.
Any number that we want to know the answers to can be added to this by performing calculations, either by adding or by directly multiplying.
Q.1: Solve 113+113+113+113+113?
Ans: We can solve this by addition as it is given in addition so,
113+113+113+113+113= 565.
The easiest way to solve this is,
113 \times 5=565
Q.2: What is the answer if we add 113 eight times?
Ans: We can solve this by adding the number 113 eight times or simply multiplying 113 by eight and we get the correct answer with the easiest method. So the answer is,
113 \times 8=904
Q.3: Solve (113 \times 2)+(113 \times 6)=?
Ans: To solve this we simply multiply 113 with two and six rather than adding it two times and then six times. The solution is,
113\times 2 = 226
113\times 6 = 678
Simply add this both to get the final answer,
(226) + (678) = 904.
Q.4: The cost of 1 kilo of sugar is 113 Rs. Find the cost of 7 kilos of sugar.
Ans: Here they gave the cost of 1 kilo of sugar and we need to find out the cost of 7 kilos of sugar, for this, we simply multiply the cost of 1 kilo of sugar by 7
Cost of 7 kilos sugar = (cost of 1-kilo sugar) ✖ (7)
= (113 Rs.) ✖ (7)
= 791 Rs.
Q.5: Calculate nine times 113.
Ans: We simply multiply 113 with 9 to get the answer,
113\times 9 = 1017
What is the 113 times table
Continuously adding 113 will provide the 113 times table,
113 |
113 + 113 = 226 |
113 + 113 + 113 = 339 |
113 + 113 + 113 + 113 = 452 |
113 + 113 + 113 + 113 + 113 = 565 |
113 + 113 + 113 + 113 + 113 + 113 = 678 |
113 + 113 + 113 + 113 + 113 + 113 + 113 = 791 |
113 + 113 + 113 + 113 + 113 + 113 + 113 + 113 = 904 |
113 + 113 + 113 + 113 + 113 + 113 + 113 + 113 + 113 = 1017 |
113 + 113 + 113 + 113 + 113 + 113 + 113 + 113 + 113 + 113 = 1130 |
Multiplication Table of 113 is given below,
113 | x | 1 | 113 |
113 | x | 2 | 226 |
113 | x | 3 | 339 |
113 | x | 4 | 452 |
113 | x | 5 | 565 |
113 | x | 6 | 678 |
113 | x | 7 | 791 |
113 | x | 8 | 904 |
113 | x | 9 | 1017 |
113 | x | 10 | 1130 |
Only one and the number itself (i.e 113) can divide the number 113.
113 has an irrational square root, and it is equal to 10.63.
Because 113 lies between the perfect squares of 100 and 121, we know that its square root falls between 10 and 11. And the square root of 113 is equal to 10.63
The first 10 multiples of 113 are 113, 226, 339, 452, 565, 678, 791, 904, 1017 and 1130.
A square of 113 means multiply 113 with 113,
113 \times 113 = 12769 .
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