Solve for x: Sol. 3x/7 + 2/7 + 4(x + 1)/5 = 2/3(2x + 1) -Maths 9th

Solve for x: Sol. 3x/7 + 2/7 + 4(x + 1)/5 = 2/3(2x + 1) -Maths 9th
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Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) |(2x-1)/(x-1)| gt 2` `(v) |x-6| le x^(2)-5x+9

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Solve for `x` `(i) |x+1|=4x+3` `(ii) |x+1|=|x+3|` `(iii) 7|x-2|-|x-7|=5` `(iv) ||x-1|-2|=6x+8` `(v) |2x^(2)-3x+1|=|x^(2)+x-3|`

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The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

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The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

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3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

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Last Answer : Solution: 7+3x = 0 ⇒ 3x = −7 ⇒ x = -7/3 ∴Remainder: 3(-7/3)3+7(-7/3) = -(343/9)+(-49/3) = (-343-(49)3)/9 = (-343-147)/9 = -490/9 ≠ 0 ∴7+3x is not a factor of 3x3+7x

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Last Answer : Solution: The equation 4x2–3x+7 can be written as 4x2–3x1+7x0 Since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x2–3x+7 is a polynomial in one variable

Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th

Description : Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th

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The ratio in which the line 3x + 4y = 7 divides the line joining the points (–2, 1) and (1, 2) is -Maths 9th

Description : The ratio in which the line 3x + 4y = 7 divides the line joining the points (–2, 1) and (1, 2) is -Maths 9th

Last Answer : (a) (–24, –2)Co-ordinates of the point of external division are\(\bigg(rac{m_1\,x_2-m_2\,x_1}{m_1-m_2},rac{m_1y_2-m_2y_1}{m_1-m_2}\bigg)\), i.e.,∴ Required point = \(\bigg(rac{3 imes-6-2 imes4}{3-2},rac{3 imes2-2 imes4}{3-2}\bigg)\)= \(\big(rac{-24}{1},rac{-2}{1}\big)\), i.e., (–24, –2).

Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x – 2x = (- 2) (3-x) (iii) (x – 2) (x + 1) = (x – 1) (x + 3) (iv) (x – 3) (2x + 1) = x (x + 5) (v) (2x – 1) (x – 3) = (x + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Description : Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x - 2x = (- 2) (3-x) (iii) (x - 2) (x + 1) = (x - 1) (x + 3) (iv) (x - 3) (2x + 1) = x (x + 5) (v) (2x - 1) (x - 3) = (x ... vi) x2 + 3x + 1 = (x - 2)2 (vii) (x + 2)3 = 2x(x2 - 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Last Answer : this is the correct answer!

Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

Description : Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

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Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Description : Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Last Answer : From x + y = 5, If x = 0 0 + y = 5 y = 5 Therefore (0,5) If x = 1 1 + y = 5 y =5 - 1 y = 4 Therefore (1,4) Draw a graph for this And From 3x - 2y = 0 If x = 0 3 (0) - 2y = 0 0 - ... 2y = 0 -2y = -6 y = -6/-2 y = 3 Therefore (2,3) Draw a graph for these points And the point of intersection is (2,3)

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

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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, \(-rac{1}{3}\)

Let f and g be the functions from the set of integers to the set integers defined by f(x) = 2x + 3 and g(x) = 3x + 2 Then the composition of f and g and g and f is given as (A) 6x + 7, 6x + 11 (B) 6x + 11, 6x + 7 (C) 5x + 5, 5x + 5 (D) None of the above

Description : Let f and g be the functions from the set of integers to the set integers defined by f(x) = 2x + 3 and g(x) = 3x + 2 Then the composition of f and g and g and f is given as (A) 6x + 7, 6x + 11 (B) 6x + 11, 6x + 7 (C) 5x + 5, 5x + 5 (D) None of the above

Last Answer : (A) 6x + 7, 6x + 11 Explanation: fog(x)=f(g(x))=f(3x+2)=2(3x+2)+3=6x+7 gof(x)=g(f(x))=g(2x+3)=3(2x+3)+2=6x+11

The linear equation 2x – 5y = 7 has -Maths 9th

Description : The linear equation 2x – 5y = 7 has -Maths 9th

Last Answer : (c) In the given equation 2x – 5y = 7, for every value of x, we get a corresponding value of y and vice-versa. Therefore, the linear equation has infinitely many solutions.

The linear equation 2x – 5y = 7 has -Maths 9th

Description : The linear equation 2x – 5y = 7 has -Maths 9th

Last Answer : (c) In the given equation 2x – 5y = 7, for every value of x, we get a corresponding value of y and vice-versa. Therefore, the linear equation has infinitely many solutions.

Determine the ratio in which 2x +3y – 30 = 0 divides the join of A(3, 4) and B(7, 8) and at what point? -Maths 9th

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Last Answer : Let A(1, 2) and B(11, 9) be the given points. Let the points of trisection be P and Q. Then,AP = PQ = QB = k (say)⇒ AQ = AP + PQ = 2k and PB = PQ + QB = 2k ∴ AP : PB = k : 2k = 1 : 2 and AQ ... two points of trisection are \(\big(rac{13}{3},rac{13}{3}\big)\) and \(\big(rac{23}{3},rac{20}{3}\big)\).

p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

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Last Answer : p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x

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Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

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Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

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If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

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If – x^2 + 3x + 4 > 0, then which of the following is correct? -Maths 9th

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Which of the following values of x do not satisfy the inequality x^2 – 3x + 2 > 0 at all. -Maths 9th

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If x is real and x^2 + 3x + 2 > 0, x^2 – 3x – 4 < 0, then which of the following is correct? -Maths 9th

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The set of values of x for which the inequalities x^2 – 3x – 10 < 0, 10x – x^2 – 16 > 0 hold simultaneously is -Maths 9th

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The equation whose roots are twice the roots of the equation x^2 – 3x + 3 = 0 is -Maths 9th

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For what value of m will the expression 3x^3 + mx^2 + 4x – 4m be divisible by x + 2 ? -Maths 9th

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Last Answer : f(x) = 3x3 + mx2 + 4x – 4m f(x) is divisible by (x + 2) if f(–2) = 0 Now f(–2) = 3(–2)3 + m(–2)2 + 4(–2) – 4m = – 24 + 4m – ... ; 4m = – 32 ≠ 0 ∴ No such value of m exists for which (x + 2) is a factor of the given expression

For what value of k, will the expression (3x^3 – kx^2 + 4x + 16) be divisible by (x – k/2) ? -Maths 9th

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Last Answer : Given f(x) = 3x³ - kx² + 4x + 16. Since (x - k/2) is a factor of polynomial. This means x = k/2 is the zero of the given polynomial. ⇒ f(k/2) = 3(k/2)³ - k(k/2)² + 4(k/2) + 16 ⇒ 0 ... - 4k + 32) + 4(k² - 4k + 32) ⇒ 0 = (k + 4)(k² - 4k + 32) ⇒ k = -4.

If the polynomials ax^3 + 4x^2 + 3x – 4 and x^3 – 4x + a leave the same remainder when divided by (x – 3), the value of a is : -Maths 9th

Description : If the polynomials ax^3 + 4x^2 + 3x – 4 and x^3 – 4x + a leave the same remainder when divided by (x – 3), the value of a is : -Maths 9th

Last Answer : Given ax^3 + 4x^2 + 3x - 4 and x^3 - 4x + a leave the same remainder when divided by x - 3. Let p(x) = ax^3 + 4x^2 + 3x - 4 and g(x) = x^3 - 4x + a By remainder theorem, if f(x) is divided by (x − a) then ... 4 27a+41 g(3)=27-4(3)+a 15+a f(3)=G(3) 27a+41=15+a 26a=15-41 a=15-41/26 a=-26/26 a=-1