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Prove that irrational
Prove that √5 is irrational.
Please click on below for answer. exercise 1.3.1
Please click on below for answer.
exercise 1.3.1
See lessFind a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.
Let the polynomial be $P(x) = ax^2 + bx + c$ Given Sum of Zeroes = -3 So, $-\frac{b}{a} = -3$ assuming a = 1 then, $-\frac{b}{1} = -3$ $\Rightarrow -b = -3$ $\Rightarrow b = 3$ Given Product of Zeroes = 2 So, $\frac{c}{a} = 2$ assuming a = 1 then, $\frac{c}{1} = 2$ $\Rightarrow c = 2$ Now a =Read more
Let the polynomial be
Given Sum of Zeroes = -3
So,
assuming a = 1 then,
Given Product of Zeroes = 2
So,
assuming a = 1 then,
Now a = 1, b = 3 and c = 2
Hence the required quadratic polynomial =
=
=
See lessShow that 3√2 is irrational.
Show that 5 – √3 is irrational.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,9m+1or9m+8
Consider the numbers , where n is a natural number. Check whether there is any value of n for which ends with the digit zero.
5. Check whether 6^n can end with the digit 0 for any natural number n.
Find the LCM and HCF of the following integers by applying the prime factorisation method of 12,15 and 21
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Sol:- Maximum number of columns = HCF(32, 616) By Euclid's division method, 616 = 32 × 19 + 8 32 = 8 × 4 + 0 ∴ HCF(32, 616) = 8 Maximum number of columns = 8
Sol:-
Maximum number of columns = HCF(32, 616)
By Euclid’s division method,
616 = 32 × 19 + 8
32 = 8 × 4 + 0
∴ HCF(32, 616) = 8
Maximum number of columns = 8
See lessExplain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers
Given that HCF (306, 657) = 9, find LCM (306, 657)
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Find the zeroes of the polynomial x² – 3 and verify the relationship between the zeroes and the coefficients.
Recall the identity a² – b² =(a – b)(a + b). Using it,we can write: x²-3=(x-√3)(x+√3) So the value of x²-3 is zero when x=√3 or x=-√3 Therefore, the zeroes of x²-3 are √3 and -√3 Sum of zeroes =√3-√3=-3/1=-(Coefficient of x)/(Coefficient of x²) Product of zeroes =(√3)(-√3)=(-3)/1 =(Constant tRead more
Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.
Sol:- Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 6. Since 0 ≤ r < 6, the possible remainders are 0, 1, 2 , 3, 4 and 5. That is, a can be 6q, or 6q + 1, or 6q + 2, or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient. However, siRead more
Sol:-
Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 6.
Since 0 ≤ r < 6, the possible remainders are 0, 1, 2 , 3, 4 and 5.
That is, a can be 6q, or 6q + 1, or 6q + 2, or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.
However, since a is odd, a cannot be 6q or 6q + 2 or 6q + 4 (since they are divisible by 2). Therefore, any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.
See lessA sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?
Sol:- This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least. Now, let us use Euclid’s algoRead more
Sol:-
This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least.
Now, let us use Euclid’s algorithm to find their HCF. We have :
420= 130×3+30
130= 30×4+10
30= 10×3+0
So, the HCF of 420 and 130 is 10.
See lessTherefore, the sweetseller can make stacks of 10 for both kinds of barfi.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Sol:- Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4. Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3. That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient. However, since a is odd, a cannot be 4Read more
Sol:-
Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4.
Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.
That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.
However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2). Therefore, any odd integer is of the form 4q + 1 or 4q + 3.
See lessFind the LCM and HCF of 6 and 20 by the prime factorisation method.
Sol:-
Sol:-
See lessFind the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Sol:- The prime factorisation of 96 and 404 gives : 96 = $2^5$ × 3, 404 = $2^2$ × 101 Therefore, the HCF of these two integers is $2^2$ = 4. Also,
Sol:-
The prime factorisation of 96 and 404 gives :
96 = × 3,
404 = × 101
Therefore, the HCF of these two integers is = 4.
Also,
See lessFind the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Sol:- We can write 6 , 72 , 120 as below 6 = 2 × 3 72 = $2^3$× $3^2$ (72 = 2×2×2×3×3) 120 = $2^3$×3×5 (120 = 2×2×2×3×5) Here, $2^1$ and $3^1$ are the smallest powers of the common factors 2 and 3 respectively. So, HCF(6,72,120)= $2^1$ × $3^1$ = 2×3 = 6 $2^3$, $3^2$ and $5^1$ are the greatest powersRead more
Sol:-
We can write 6 , 72 , 120 as below
6 = 2 × 3
72 = × (72 = 2×2×2×3×3)
120 = ×3×5 (120 = 2×2×2×3×5)
Here, and are the smallest powers of the common factors 2 and 3 respectively.
So, HCF(6,72,120)= × = 2×3 = 6
, and are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So, LCM(6,72,120)= × × =360
See lessUse Euclid’s division algorithm to find the HCF of 867 and 255
867 can be written in multiples of 255 as below. 867 = 255 * 3 + 102 255 = 102 * 2 + 51 102 = 51 * 2 + 0 As you can see the remainder has become zero now and the highest common factor is 51.
867 can be written in multiples of 255 as below.
867 = 255 * 3 + 102
255 = 102 * 2 + 51
102 = 51 * 2 + 0
As you can see the remainder has become zero now and the highest common factor is 51.
See lessUse Euclid’s division algorithm to find the HCF of 135 and 225
225 = 135*1+90 135=90*1+45 90=45*2+0 Therefore HCF is 45
225 = 135*1+90
135=90*1+45
90=45*2+0
Therefore HCF is 45
See lessUse Euclid’s division algorithm to find the HCF of 196 and 38220
According to Euclid's theorem x = y*q + r. 38220 = 196×195 + 0 So the HCF is 196.
According to Euclid’s theorem x = y*q + r.
38220 = 196×195 + 0
So the HCF is 196.
See lessShow that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.
Sol:- Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1. If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positivRead more
Sol:-
Let a be any positive integer and b = 2. Then, by Euclid’s algorithm,
a = 2q + r
for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1.
If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1.
See lessUse Euclid’s algorithm to find the HCF of 4052 and 12576.
Sol:- Since 12576 > 4052 so, we will divide 12576 by 4052 and follow below steps till we will get 0 as a remainder. 12576 = 4052×3 + 420 4052 = 420×9 + 272 420 = 272×1 + 148 272 = 148×1 + 124 148 = 124×1 + 24 124 = 24×5 + 4 24 = 4×6 +0 The remainder has now become zero, so our procedure stops herRead more
Sol:- Since 12576 > 4052 so, we will divide 12576 by 4052 and follow below steps till we will get 0 as a remainder.
12576 = 4052×3 + 420
4052 = 420×9 + 272
420 = 272×1 + 148
272 = 148×1 + 124
148 = 124×1 + 24
124 = 24×5 + 4
24 = 4×6 +0
The remainder has now become zero, so our procedure stops here. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.
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