4s² - 4s +1 = 4s² - 2s - 2s + 1 = 2s(2s - 1) - 1(2s - 1) = (2s - 1)(2s - 1) The value of 4s² - 4s + 1 is zero if 2s - 1 = 0 ⇒ s = 1/2 Therefore, the zeroes of 4s² - 4s + 1 are 1/2 and 1/2. Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Cofficient of s)/(Cofficient of s²) Product of zeroes = 1/2 x 1Read more
4s² – 4s +1
= 4s² – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
The value of 4s² – 4s + 1 is zero if 2s – 1 = 0 ⇒ s = 1/2
Therefore, the zeroes of 4s² – 4s + 1 are 1/2 and 1/2.
Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Cofficient of s)/(Cofficient of s²)
Product of zeroes = 1/2 x 1/2 = 1/4 = 1/4 = (Constant term)/Coefficient of s²
29/343 343 = 7x7x7 = 7³ Since the denominator is not in the form 2^m x 5^n, and it has 7 as its factor, the decimal expansion of 29/343 is non-terminating repeating. See this 👇
29/343
343 = 7x7x7 = 7³
Since the denominator is not in the form 2^m x 5^n, and it has 7 as its factor, the decimal expansion of 29/343 is non-terminating repeating.
129/2²5⁷7⁵ Since the denominator is not of the form 2^m x 5^n, and it also has 7 as its factor, the decimal expansion of 129/2²5⁷7⁵ non-terminating repeating. Explanation Video 😊👇
129/2²5⁷7⁵
Since the denominator is not of the form 2^m x 5^n, and it also has 7 as its factor, the decimal expansion of 129/2²5⁷7⁵ non-terminating repeating.
Let 6 + √2 be rational. Therefore, we can find two co-prime integers a, b (b 0) such that 6 + √2 = a/b ⇒ √2 = a/b - 6 Since a and b are integers,a/b - 6 is also rational and hence, √2 should be rational. This contradicts the fact that √2 is irrational. Therefore, our assumption is false and hence, 6Read more
Let 6 + √2 be rational.
Therefore, we can find two co-prime integers a, b (b 0) such that
6 + √2 = a/b
⇒ √2 = a/b – 6
Since a and b are integers,a/b – 6 is also rational and hence, √2 should be rational. This
contradicts the fact that √2 is irrational. Therefore, our assumption is false and hence,
6 + √2 is irrational.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 4s² – 4s + 1.
4s² - 4s +1 = 4s² - 2s - 2s + 1 = 2s(2s - 1) - 1(2s - 1) = (2s - 1)(2s - 1) The value of 4s² - 4s + 1 is zero if 2s - 1 = 0 ⇒ s = 1/2 Therefore, the zeroes of 4s² - 4s + 1 are 1/2 and 1/2. Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Cofficient of s)/(Cofficient of s²) Product of zeroes = 1/2 x 1Read more
4s² – 4s +1
= 4s² – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
The value of 4s² – 4s + 1 is zero if 2s – 1 = 0 ⇒ s = 1/2
Therefore, the zeroes of 4s² – 4s + 1 are 1/2 and 1/2.
Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Cofficient of s)/(Cofficient of s²)
Product of zeroes = 1/2 x 1/2 = 1/4 = 1/4 = (Constant term)/Coefficient of s²
Explanation 🤠
See lessWithout actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 29 / 343
29/343 343 = 7x7x7 = 7³ Since the denominator is not in the form 2^m x 5^n, and it has 7 as its factor, the decimal expansion of 29/343 is non-terminating repeating. See this 👇
29/343
343 = 7x7x7 = 7³
Since the denominator is not in the form 2^m x 5^n, and it has 7 as its factor, the decimal expansion of 29/343 is non-terminating repeating.
See this 👇
See lessWithout actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 23 / 2³5²
23/2³5² Denominator = 2³5² The denominator is of the form 2^m x 5^n. Hence, the decimal expansion of 23/2³5² is terminating. Explanation 👌
23/2³5²
Denominator = 2³5²
The denominator is of the form 2^m x 5^n.
Hence, the decimal expansion of 23/2³5² is terminating.
Explanation 👌
See lessWithout actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 129 / 2² 5⁷ 7⁵
129/2²5⁷7⁵ Since the denominator is not of the form 2^m x 5^n, and it also has 7 as its factor, the decimal expansion of 129/2²5⁷7⁵ non-terminating repeating. Explanation Video 😊👇
129/2²5⁷7⁵
Since the denominator is not of the form 2^m x 5^n, and it also has 7 as its factor, the decimal expansion of 129/2²5⁷7⁵ non-terminating repeating.
Explanation Video 😊👇
See lessProve that the following is irrational: 6+√2
Let 6 + √2 be rational. Therefore, we can find two co-prime integers a, b (b 0) such that 6 + √2 = a/b ⇒ √2 = a/b - 6 Since a and b are integers,a/b - 6 is also rational and hence, √2 should be rational. This contradicts the fact that √2 is irrational. Therefore, our assumption is false and hence, 6Read more
Let 6 + √2 be rational.
Therefore, we can find two co-prime integers a, b (b 0) such that
6 + √2 = a/b
⇒ √2 = a/b – 6
Since a and b are integers,a/b – 6 is also rational and hence, √2 should be rational. This
contradicts the fact that √2 is irrational. Therefore, our assumption is false and hence,
6 + √2 is irrational.
See here👇
See less