Find the value of logxx + logxx^3 + logxx^5 + ........ + log xx^(2n-1). -Maths 9th

Find the value of logxx + logxx^3 + logxx^5 + ........ + log xx^(2n-1). -Maths 9th
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If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : Let l+m−2nlogx​=m+n−2llogy​=n+l−2mlogz​=k(say) So, we get logx=k(l+m−2n) ....... (i) logy=k(m+n−2l) ....... (ii) logz=k(n+l−2m) ....... (iii) ∴logx+logy+logz=k(l+m−2n)+k(m+n−2l)+k(n+l−2m) ⇒logx+logy+logz=kl+km−2kn+km+kn−2kl+kn+kl−2km ⇒log(xyz)=0 ⇒logxyz=log1 ⇒xyz=1

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Last Answer : (b) 1Let \(rac{ ext{log}\,x}{l+m-2n}\) = \(rac{ ext{log}\,y}{m+n-2l}\) = \(rac{ ext{log}\,z}{n+l-2m}\) = k. Thenlog x = k(l + m – 2n), log y = k(m + n – 2l); log z = k(n + l – 2m) ⇒ log x + log y + log z = k(l + m – 2n) + k(m + n – 2l) + k(n + l – 2m)⇒ log(xyz) = 0 ⇒ log(xyz) = log 1 ⇒ xyz = 1.

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Description : If 1/3 log3 M + 3 log3 N = 1 + log 0.008 5, then -Maths 9th

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Description : If (log x)/(a^2+ab+b^2) = (log y)/(b^2+bc+c^2) = (log z)/(c^2+ca+a^2), then x^(a-b). y^(b-c). z^(c-a) = -Maths 9th

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Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

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Description : Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

Last Answer : Let p(x) =3x3 – 4x2 + 7x – 5 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3)- 5 = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

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Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

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Description : If a+b+c=5 and ab+bc+ca=10 find the value of a^3+b^3+c^3-3abc -Maths 9th

Last Answer : We know , a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca) now , a + b + c = 5 ab + bc + ca = 10 (a + b + c)² = a² + b² + c² +2(ab + bc+ca) (5)² -2 10 = a² + b² + c² a² + b² + c² = ... )(a² + b² + c² -ab- bc-ca) =( 5)( 5 - 10) = 5 (-5) = -25 Hope this will help u..... by :- RAXTAR.....

If the mode of the data 5, 8, 4,5,5,8, 4, 7, 8, x is 5, then find the value of x. -Maths 9th

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Last Answer : (8+5+2+x+6+5)/6=6 (26+x)=6×6 26+x=36 X=36-26 X=10

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Description : Find the value of x, if log2 (5.2^x + 1), log4(2^(1–x) + 1) and 1 are in A.P. -Maths 9th

Last Answer : (b) 1 - log25 Given, log2 (5.2x + 1), log4 (21- x + 1), 1 are in A.P. ⇒ log2 (5.2x + 1) + 1 = 2 log4 (21 - x + 1) ⇒ log2 (5.2x + 1) + log22 = 2 log22 (21-x + 1)⇒ log2 (5.2x + 1).2 = 2 x \(rac12\) ... a=-rac{1}{2}\big)\)⇒ log 2x = log \(rac{2}{5}\)⇒ x log2 2 = log2 2 - log2 5 ⇒ \(x\) = 1 - log2 5.

If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Description : If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Last Answer : (d) \(-rac{1}{3}\)\(5^{{3x^2}log_{10}2}\) = 2\(\big(x+rac{1}{2}\big)\)log10 25⇒ \(5^{{3x^2}log_{10}2}\) = 2\(\big(rac{2x+1}{2}\big)\) x log10 5 = 2(2x+1)log10 5⇒ \(5^{{3x^2}log_{10}2}\) = 2(2x+1)log2 5. ... aloga x = x]⇒ 3x2 = 2x + 1 ⇒ 3x2 - 2x - 1 = 0 ⇒ (x - 1) (3x + 1) = 0⇒ x = 1, \(-rac{1}{3}\)

The minimum value of the sum of real numbers a^(–5), a^(–4), 3a^(–3), 1, a^8 and a^10 with a > 0 is -Maths 9th

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If A lies in the third quadrant and 3 tan A – 4 = 0, then what is the value of 5 sin 2 A + 3 sin A + 4 cos A? -Maths 9th

Description : If A lies in the third quadrant and 3 tan A – 4 = 0, then what is the value of 5 sin 2 A + 3 sin A + 4 cos A? -Maths 9th

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For what value of m the ratio of the roots of the equation 12x^2 – mx + 5 = 0 is 3 : 2 ? -Maths 9th

Description : For what value of m the ratio of the roots of the equation 12x^2 – mx + 5 = 0 is 3 : 2 ? -Maths 9th

Last Answer : Given equation: 12x2+mx+5=0 The roots are in ratio 3:2 Hence,let roots of the equations are 3α and 2α. Applying condition for sum and product of the roots, 3α+2α=−12m​ and 3α×2α=125​⇒α2=725​⇒α=±62​5​​And m=−60α ⇒m=±510​Hence, A is the correct option.

If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Description : Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Last Answer : Let C(x, y) be the centre of the circle passing through the points P(6, -6), Q(3, -7) and R(3, 3) Then, PC = QC = RC(Being radius of the same circle) PC2 = QC2 ⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 ⇒ x2 - ... i), we get 3\(x\) + (-2) - 7 = 0 ⇒ 3\(x\) = 9 ⇒ \(x\) = 3 ∴ The centre is (3, -2).

If points A (4, 3) and B (x, 5) are on the circle with centre O (2, 3), find the value of x. -Maths 9th

Description : If points A (4, 3) and B (x, 5) are on the circle with centre O (2, 3), find the value of x. -Maths 9th

Last Answer : Since A and B lie on the circle having centre O. Therefore, OA = OB root(4−2)2+(3−3)2​=(root(x−2)2+(5−3)2​2=root(x−2)2+4​(x−2)2+4=4 (x−2)2=0 x−2=0 x=2 hope it helps thank you

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Description : Find the value of x and y if l is parallel to m -Maths 9th

Last Answer : the value should be zero...according to me as I think!!

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Description : Find the decimal expansion of 7/22 .Explain why it is taken as the approximate value of pi -Maths 9th

Last Answer : 3.1564267467648582474747477647908765434567896542345678765431222245775583585.................................. or 22/7 is the value f pi

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Description : What's pie's value -Maths 9th

Last Answer : The number π is a mathematical constant, the ratio of a circle's circumference to its diameter, commonly approximated as 3.14159. It has been represented by the Greek letter 'π' since the mid-18th century, though it is also sometimes spelled