Solve for x: 5(4x + 3) = 3(x -2) -Maths 9th

Description : In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

Last Answer : when the work is done for 6 hours x=6 y=4(6)+13 y=24+13 y=37 the labourer gets Rs.37 if he works for 6hrs

If loge (x^2 – 16) < loge (4x – 11), then -Maths 9th

Description : If loge (x^2 – 16) < loge (4x – 11), then -Maths 9th

Last Answer : answer:

If sin^4x + sin^2x = 1 then what is 1 are the value of cot^4 x + cot^2 x? -Maths 9th

Description : If sin^4x + sin^2x = 1 then what is 1 are the value of cot^4 x + cot^2 x? -Maths 9th

Last Answer : answer:

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

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

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

If x^2 – 4x + 1 = 0, then what is the value of x^3 + 1/x^3? -Maths 9th

Description : If x^2 – 4x + 1 = 0, then what is the value of x^3 + 1/x^3? -Maths 9th

Last Answer : Hope its clearrrrrrr!!!!!!!!!!!!!!!!!!!

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

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

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

When x^3 + 2x^2 + 4x + b is divided by (x + 1), the quotient is x^2 + ax + 3 and the remainder -Maths 9th

Description : When x^3 + 2x^2 + 4x + b is divided by (x + 1), the quotient is x^2 + ax + 3 and the remainder -Maths 9th

Last Answer : answer:

If 4x^2 – 6x + m is divisible by x – 3, which one of the following is the greatest divisor of m ? -Maths 9th

Description : If 4x^2 – 6x + m is divisible by x – 3, which one of the following is the greatest divisor of m ? -Maths 9th

Last Answer : answer:

Write two solutions of the equation 4x -5 y = 15. -Maths 9th

Description : Write two solutions of the equation 4x -5 y = 15. -Maths 9th

Last Answer : Solution :-

Find any four solutions of the equation 4x+3y=12. -Maths 9th

Description : Find any four solutions of the equation 4x+3y=12. -Maths 9th

Last Answer : Given equation is 4x + 3y =12 On putting x = 0 in Eq. (i), we get 4(0) +3y =12 ⇒ 3y =12 ⇒ y = 12 / 3 = 4 So, (0, 4) is a solution of given equation. On putting y = 0 in Eq. (i), we ... given equation. Hence the four solutions of given equation are (0, 4), (3, 0), (1, 8 / 3) and (2,4 / 3).

Find any four solutions of the equation 4x+3y=12. -Maths 9th

Description : Find any four solutions of the equation 4x+3y=12. -Maths 9th

Last Answer : Given equation is 4x + 3y =12 On putting x = 0 in Eq. (i), we get 4(0) +3y =12 ⇒ 3y =12 ⇒ y = 12 / 3 = 4 So, (0, 4) is a solution of given equation. On putting y = 0 in Eq. (i), we ... given equation. Hence the four solutions of given equation are (0, 4), (3, 0), (1, 8 / 3) and (2,4 / 3).

factorise the following 4x^2+20x+25 -Maths 9th

Description : factorise the following 4x^2+20x+25 -Maths 9th

Last Answer : NEED ANSWER

factorise the following 4x^2+20x+25 -Maths 9th

Description : factorise the following 4x^2+20x+25 -Maths 9th

Last Answer : 4x^2+20x+25 4x^2+10x+10x+25 2x(2x+5)+5(2x+5) (2x+5)(2x+5) 2x^2+2(2x)(5)+25 2x^2+20x+25

Show that 2x+1 is a factor of polynomial 2x(cube) - 11x(square) - 4x + 1. -Maths 9th

Description : Show that 2x+1 is a factor of polynomial 2x(cube) - 11x(square) - 4x + 1. -Maths 9th

Last Answer : Solution :-

Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th

Description : Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th

Last Answer : Solution :-

Find the coordinate where the linear equation 4x - 23 y = 7 meets at y-axis. -Maths 9th

Description : Find the coordinate where the linear equation 4x - 23 y = 7 meets at y-axis. -Maths 9th

Last Answer : 4x-2=-7*3y 4x+21y=2 The equation meets y axis when x=0 4.0+21y=2 y=21/2 Hence , the equation meets y-axis at (0,21/2)

The angles of a quadrilateral are 4x degree,7x degree, 15x degree, 10x degree.find the smallest and largest of the quadrilateral. -Maths 9th

Description : The angles of a quadrilateral are 4x degree,7x degree, 15x degree, 10x degree.find the smallest and largest of the quadrilateral. -Maths 9th

Last Answer : Solution :-

The solution of 4x^2 + 4x + 1 > 0 is -Maths 9th

Description : The solution of 4x^2 + 4x + 1 > 0 is -Maths 9th

Last Answer : answer:

What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Description : What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Last Answer : (b) (5, 6)Let the foot of the perpendicular be M(x1, y1) Slope of line AB, i.e., y = -x + 11 = -1 Slope of line PM = \(rac{y_1-3}{x_1-2}\)Now, PM ⊥ AB⇒ \(\bigg(rac{y_1-3}{x_1-2}\bigg)\) x - ... get 2x1 = 10 ⇒ x1 = 5 Putting x1 in (ii), we get y1 = 6. ∴ Required foot of the perpendicular M is (5, 6).

The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th

Description : The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th

Last Answer : (c) 25x + 25y - 4 = 0 The point of intersection of the given lines can be obtained by solving the equations of the two lines simultaneously. 100x + 50y = 1 ...(i) 75x + 25y = -3 ... = \(rac{4}{25}\)∴ Eqn of required line: x + y = \(rac{4}{25}\) ⇒ 25 + 25y - 4 = 0.

Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

Description : Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

Last Answer : answer:

Find the following products: (i) (3x + 2y) (9X2 – 6xy + Ay2) (ii) (4x – 5y) (16x2 + 20xy + 25y2) (iii) (7p4 + q) (49p8 – 7p4q + q2) -Maths 9th

Description : Find the following products: (i) (3x + 2y) (9X2 – 6xy + Ay2) (ii) (4x – 5y) (16x2 + 20xy + 25y2) (iii) (7p4 + q) (49p8 – 7p4q + q2) -Maths 9th

Last Answer : answer:

Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12. -Maths 9th

Description : Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12. -Maths 9th

Last Answer : answer:

Solve graphically 4x-3y+4=0, 4x+3y-20=0 -Maths 10th

Description : Solve graphically 4x-3y+4=0, 4x+3y-20=0 -Maths 10th

Last Answer : 4x−3y+4=0 ⟹x=4−4+3y Let, y=4, then x=4−4+3(4) =2 Similarly, when y=0,x=−1 4x+3y−20=0 ⟹x=420−3y Let, y=4, then x=420−3(4) =2 Similarly, when y=0,x=5 Plotting the above points on the graph, ... point i.e(2,4) Area of the region bounded by these lines and x-axis =21 6units 4 units =12 sq. units.

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

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

Last Answer : Solution :-

Solve (|x – 1| – 3) (|x + 2| – 5) < 0. Then -Maths 9th

Description : Solve (|x – 1| – 3) (|x + 2| – 5) < 0. Then -Maths 9th

Last Answer : Case 1. |x-1|-3>0 |x+2|–53, |x+2|

Solve the equation 2x + 1 = x -3, and represent the solution(s) on (i) the number line. (ii) the Cartesian plane. -Maths 9th

Description : Solve the equation 2x + 1 = x -3, and represent the solution(s) on (i) the number line. (ii) the Cartesian plane. -Maths 9th

Last Answer : Solution :-

Solve for x if a > 0 and 2 logx a + logax a + 3 loga2x a = 0 -Maths 9th

Description : Solve for x if a > 0 and 2 logx a + logax a + 3 loga2x a = 0 -Maths 9th

Last Answer : (d) \(a^{-rac{4}{3}}\)Since logaxa = \(rac{1}{log_aax}\) = \(rac{1}{log_aa+log_ax}\) = \(rac{1}{1+log_ax}\) andloga2x a = \(rac{1}{log_aa^2x}\) = \(rac{1}{log_aa^2+log_{ax}x}\) = \(rac{1}{2log_aa+log_{ax}x}\)= \( ... 2}}\)When t = \(-rac{4}{3}\), logax = \(-rac{4}{3}\) ⇒ \(x\) = \(a^{-rac{4}{3}}\)

Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Description : Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Last Answer : (b) \(\bigg(rac{10}{3},rac{20}{3}\bigg)\). (+ 10, 20) log100 |x+y| = \(rac{1}{2}\) ⇒ |x + y| = 100\(^{rac{1}{2}}\)⇒ |x + y| = 10 as (-10 is inadmissible) ...(i) log10y - log10| x | = log1004⇒ log10 ... x < 0, then x = 10.∴ If x = \(rac{10}{3}\), then y = \(rac{20}{3}\) and if x = 10, y = 20.

Solve x^2 – 5x + 4 > 0. -Maths 9th

Description : Solve x^2 – 5x + 4 > 0. -Maths 9th

Last Answer : answer:

Solve : 4^(1 + x) + 4^(1 – x) = 10 for x. -Maths 9th

Description : Solve : 4^(1 + x) + 4^(1 – x) = 10 for x. -Maths 9th

Last Answer : answer:

Solve : (x + 1) (x + 2) (x + 3) (x + 4) + 1 = 0 -Maths 9th

Description : Solve : (x + 1) (x + 2) (x + 3) (x + 4) + 1 = 0 -Maths 9th

Last Answer : answer:

Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

Solve for x : log10 [log2 (log39)] = x -Maths 9th

Description : Solve for x : log10 [log2 (log39)] = x -Maths 9th

Last Answer : answer:

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|`

Description : 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|`

Last Answer : 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|`

Use the graph to solve the system of linear equations x - y=4 4x + y=1?

Description : Use the graph to solve the system of linear equations x - y=4 4x + y=1?

Last Answer : x=4+y4x+y=1 = 4(4+y)+y=116+4y+y=116+5y=15y=-15y=-3x=4+y = x=4+(-3)x=1

how do i solve 4x-9=k for x?

Description : how do i solve 4x-9=k for x?

Last Answer : 4x-9=k4x=k+9Divide by 4 to isolate the xx=(1/4)k+(9/4)x=1/4 k+ 9/4

Solve the following equations (where `[*]` dentoes greatest integer function and `{*}` represent fractional part function) `(i) 2[x]+3[x]=4x-1` `(ii)

Description : Solve the following equations (where `[*]` dentoes greatest integer function and `{*}` represent fractional part function) `(i) 2[x]+3[x]=4x-1` `(ii)

Last Answer : Solve the following equations (where `[*]` dentoes greatest integer function and `{*}` represent fractional part function) ... }` `(iii) [x]+2{-x}=3x`

Solve the following equations : `(i) log_(x)(4x-3)=2` `(ii) log_2)(x-1)+log_(2)(x-3)=3` `(iii) log_(2)(log_(8)(x^(2)-1))=0` `(iv) 4^(log_(2)x)-2x-3=0`

Description : Solve the following equations : `(i) log_(x)(4x-3)=2` `(ii) log_2)(x-1)+log_(2)(x-3)=3` `(iii) log_(2)(log_(8)(x^(2)-1))=0` `(iv) 4^(log_(2)x)-2x-3=0`

Last Answer : Solve the following equations : `(i) log_(x)(4x-3)=2` `(ii) log_2)(x-1)+log_(2)(x-3)=3` `(iii) log_(2)(log_(8)(x^(2)-1))=0` `(iv) 4^(log_(2)x)-2x-3=0`

Solve for x: 5(4x + 3) = 3(x -2) -Maths 9th

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