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Answered on 17 Apr Learn Euclid's Division Lemma
Nazia Khanum
To find aa, we can use the relationship between the highest common factor (HCF) and the least common multiple (LCM).
Given: HCF(6,a)=2HCF(6,a)=2 LCM(6,a)=60LCM(6,a)=60
We know that the product of the HCF and LCM of two numbers is equal to the product of the numbers themselves. So, we have:
HCF(6,a)×LCM(6,a)=6×aHCF(6,a)×LCM(6,a)=6×a
Substituting the given values:
2×60=6×a2×60=6×a 120=6a120=6a
Now, we solve for aa:
a=1206a=6120 a=20a=20
Therefore, a=20a=20.
Answered on 17 Apr Learn Euclid's Division Lemma
Nazia Khanum
To find the size of tiles required and how many tiles are needed, we first need to determine the largest possible size of the square marble tiles that can fit evenly into the room's dimensions.
Given:
To find the largest possible size of the square tiles, we need to find the greatest common divisor (GCD) of the dimensions (length and width) of the room.
GCD(10, 7) = 1
This means that there's no integer length that evenly divides both 10 and 7. So, we cannot find a single tile size that perfectly fits the room without cutting.
However, we can approximate the largest possible tile size by using factors of the GCD, which in this case is 1. So, theoretically, we could use a tile size of 1m x 1m, but that wouldn't be practical.
In real-world scenarios, the tile size is usually chosen for convenience and aesthetics. A common approach is to use a tile size that evenly divides the room dimensions, even if it involves some cutting.
Let's say we decide to use a 0.5m x 0.5m tile size. Then, we can calculate how many tiles are needed:
For the length of the room (10m), we would need 10m / 0.5m = 20 tiles. For the width of the room (7m), we would need 7m / 0.5m = 14 tiles.
So, in total, we would need 20 tiles lengthwise and 14 tiles widthwise, resulting in 20 x 14 = 280 tiles.
However, keep in mind that some tiles will need to be cut to fit the edges of the room, especially along one of the dimensions (either length or width). The number of tiles that need to be cut will depend on the exact layout of the tiles and the dimensions of the room.
Answered on 26/11/2022 Learn Euclid's Division Lemma
Sandhya
Maths tutor with 7 years experience
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 17 Apr Learn Rational and irrational numbers
Nazia Khanum
To show that 2+22+2
is not a rational number, we'll use a proof by contradiction.
Assume that 2+22+2
is rational. That means it can be expressed as the ratio of two integers aa and bb where b≠0b=0 and aa and bb have no common factors other than 1:
2+2=ab2+2
=ba
Now, let's rearrange this equation to isolate 22
:
2=ab−22
=ba−2
2=a−2bb2
=ba−2b
Now, square both sides:
2=(a−2bb)22=(ba−2b)2
2=(a−2b)2b22=b2(a−2b)2
2b2=(a−2b)22b2=(a−2b)2
2b2=a2−4ab+4b22b2=a2−4ab+4b2
0=a2−4ab+2b20=a2−4ab+2b2
This equation represents a quadratic equation in terms of aa. Now, let's consider this equation modulo 2. This means we'll look at the remainders when dividing each term by 2.
0≡a2−4ab+2b2(mod2)0≡a2−4ab+2b2(mod2)
0≡a2(mod2)0≡a2(mod2)
Since the square of any integer is congruent to either 0 or 1 modulo 2, a2≡0(mod2)a2≡0(mod2) implies that aa itself must be even.
Let a=2ka=2k, where kk is an integer.
Now, substitute a=2ka=2k into the equation:
0=(2k)2−4(2k)b+2b20=(2k)2−4(2k)b+2b2
0=4k2−8kb+2b20=4k2−8kb+2b2
0=2(2k2−4kb+b2)0=2(2k2−4kb+b2)
Since 22 is a prime number, for 2(2k2−4kb+b2)2(2k2−4kb+b2) to be 00, the term inside the parentheses must be divisible by 22. But if 22 divides 2k2−4kb+b22k2−4kb+b2, then 22 divides each of its terms, including b2b2. This implies that bb is also even.
Now, if both aa and bb are even, then they have a common factor of 22, contradicting our initial assumption that aa and bb have no common factors other than 1.
Thus, our initial assumption that 2+22+2
is rational must be false. Therefore, 2+22+2
is irrational.
Answered on 17 Apr Learn Rational and irrational numbers
Nazia Khanum
Sure, let's consider the rational number 1221 and the irrational number 22
.
The product of 1221 and 22
is:
12×2=2221×2
=22
Here, we have a rational number (1221) multiplied by an irrational number (22
), resulting in another rational number (2222
). Therefore, this example demonstrates that the product of a rational number and an irrational number can indeed be rational.
Answered on 16 Apr Learn Rational and irrational numbers
Sadika
Sure, there are infinitely many irrational numbers between any two distinct real numbers. Here are two irrational numbers between √2 and √3
:
Both of these numbers are irrational because they cannot be expressed as a fraction of two integers and they lie between √2
and √3
.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 16 Apr Learn Rational and irrational numbers
Sadika
To prove that the 5 +3√2
Answered on 16 Apr Learn Rational and irrational numbers
Sadika
To find the smallest number divisible by both 360 and 657, we need to find the least common multiple (LCM) of these two numbers.
Answered on 16 Apr Learn Rational and irrational numbers
Sadika
Sure, let's prove that the square of any positive integer cannot be of the form 5q+25q+2 or 5q+35q+3 for any integer q.
We'll consider two cases:
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 16 Apr Learn Rational and irrational numbers
Sadika
Consider any three consecutive positive integers. Let's denote them as nn, n+1n+1, and n+2n+2.
Now, let's check the divisibility of each integer by 3:
For nn: If nn is divisible by 3, then we have one integer in the set that is divisible by 3. If not, nn leaves a remainder of either 1 or 2 when divided by 3.
For n+1n+1: If nn leaves a remainder of 0 when divided by 3, then n+1n+1 leaves a remainder of 1. If nn leaves a remainder of 1 when divided by 3, then n+1n+1 leaves a remainder of 2. And if nn leaves a remainder of 2 when divided by 3, then n+1n+1 leaves a remainder of 0. In any case, n+1n+1 is not divisible by 3.
For n+2n+2: If nn leaves a remainder of 0 when divided by 3, then n+2n+2 leaves a remainder of 2. If nn leaves a remainder of 1 when divided by 3, then n+2n+2 leaves a remainder of 0. And if nn leaves a remainder of 2 when divided by 3, then n+2n+2 leaves a remainder of 1. In any case, n+2n+2 is not divisible by 3.
So, out of the three consecutive integers nn, n+1n+1, and n+2n+2, exactly one of them is divisible by 3, as the other two always leave a remainder of either 1 or 2 when divided by 3.
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