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Answered on 24 Feb Learn Playing With Numbers
Sadika
The sum of any two even numbers is always an even number.
Let's denote two even numbers as 2n2n and 2m2m, where nn and mm are integers.
The sum of these two even numbers is: 2n+2m=2(n+m)2n+2m=2(n+m)
Since nn and mm are integers, n+mn+m is also an integer. Therefore, the sum 2(n+m)2(n+m) is an even number because it is divisible by 22. Thus, the sum of any two even numbers is an even number.
Answered on 24 Feb Learn Playing With Numbers
Sadika
To find the product of two numbers given their least common multiple (LCM) and highest common factor (HCF), you can use the relationship between the product, LCM, and HCF of two numbers.
Let the two numbers be xx and yy.
Given:
We know that the product of two numbers is equal to the product of their LCM and HCF. That is:
x×y=LCM(x,y)×HCF(x,y)x×y=LCM(x,y)×HCF(x,y)
Substitute the given values:
x×y=15×4x×y=15×4
x×y=60x×y=60
So, the product of the two numbers is 6060.
Answered on 24 Feb Learn Practical Geometry
Sadika
To construct a perpendicular to a given line segment at a point on it, you can follow these steps:
Now, you have constructed a perpendicular to the given line segment at the chosen point PP.
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Answered on 24 Feb Learn Knowing our Numbers
Sadika
The greatest three-digit number that remains the same when its digits are reversed is also a palindrome.
The greatest three-digit palindromic number is 999.
When you reverse the digits of 999, you still get 999. So, it is the greatest three-digit number that does not change when its digits are reversed.
Answered on 24 Feb Learn Knowing our Numbers
Sadika
To get a number whose digits are the reverse of 203, we need to find the difference between this number and its reverse.
The number 203, when its digits are reversed, becomes 302.
To find what must be added to 203 to get 302, we subtract 203 from 302:
302−203=99302−203=99
So, 99 must be added to 203 to get a number whose digits are the reverse of the given number.
Answered on 24 Feb Learn Ratio and Proportion
Sadika
To divide 60 in the ratio of 2 : 3, we need to find two parts such that their ratio is 2 : 3.
Let the parts be 2x and 3x, where x is the common factor.
According to the given ratio: 2x+3x=602x+3x=60
Combine like terms: 5x=605x=60
Now, solve for x: x=605x=560 x=12x=12
Now, we can find the values of the parts: Part 1=2x=2×12=24Part 1=2x=2×12=24 Part 2=3x=3×12=36Part 2=3x=3×12=36
So, the parts in the ratio of 2 : 3 are 24 and 36, respectively.
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Answered on 24 Feb Learn Basic Geometrical ideas
Sadika
A triangle has three vertices. Each vertex is a point where two sides of the triangle meet. So, regardless of the type of triangle (equilateral, isosceles, or scalene), it always has three vertices.
Answered on 24 Feb Learn Understanding elementary Shapes
Sadika
To find the fraction of a clockwise revolution that the hour hand of a clock turns through when it goes from 3 to 9, we first need to determine the total angle turned by the hour hand in a clockwise direction.
The total angle of a clock is 360 degrees. The hour hand moves through 12 hours to complete one full revolution.
From 3 to 9, the hour hand moves through 6 hours.
So, the fraction of a clockwise revolution turned through by the hour hand from 3 to 9 is:
Fraction=Number of hours passed/Total hours in a clock=6/12=1/2
Therefore, the hour hand turns through 1221 of a clockwise revolution when it goes from 3 to 9 on a clock.
Answered on 24 Feb Learn Understanding elementary Shapes
Sadika
If the hour hand of a clock starts at 12 and makes 1/2 of a revolution clockwise, it will stop at the position that is exactly opposite its starting position.
Since a full revolution of the hour hand covers 360 degrees, and it makes 1/2 of a revolution clockwise, it will cover 1/2×360∘=180∘
Therefore, the hour hand will stop at the position exactly opposite to 12, which is 6 on the clock face.
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Answered on 24 Feb Learn symmetry
Sadika
The letter "O" of the English alphabet has infinite lines of symmetry.
Any line passing through the center of the letter "O" will divide it into two mirror-image halves that are congruent. Since there are infinite lines that can pass through the center of the letter "O", there are infinite lines of symmetry for the letter "O".
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