solve the equation

24n = 120. What is n?

25c^2+10c+1

2x+5=x^2

The numbers in brackets are powers

Solve (D2-3D+2)y =x(x+4) and show that its general solution is given by y=Aex + Be2x +(x2/2) + (7x/2) + (19/4)

My work :

auxiliary eqn , m2 - 3m + 2= 0 m=1 , m=2

Reduced eqn : yn = Aex + Be2x

P.I. = 1/f(D)F(x) = (1/(D2-3D+2))(x(x+4))

P.I. = [(x2+4x) + ( (1/2)( 2- 3(2x) - 3(4) ) ) + (1/4)(92) )

P.I. = x2 + (4x/2) - (6x/4) - (5/2) + (18/4)

P.I. =(x2/2) + x + 8/4

General solution : y= Aex + Be2x + (x2/2) + x + 8/4

Could anyone point out my mistakes?

differentiate f(t)=2-(2/3)t

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find x = 2(3x×x)+(2x×)=14400

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Find what y will be when x=40 degrees 2) Sketch the curve for one positive and one negative cycle and then on your graph determine the x value when y is 9

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-4=8x+y

5x+3/4x-7≥2, can anyone solve this inequality and show the step by step method of doing it, Many thanks!!

Solve:

3x(x+4)+3(x+4)=0

what is 1cmx1cmx1cm im mm2?

3x⁴-2x³-5x²-4/x²-4

HOW CAN I SOVLE QUADRATIC SIMTENEOUS EQUATION

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please can you give me a step by step guide

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Eg. 4x2 + 3x - 8?

like i dont get how to do this problem:3y-4=2

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I need help with this please help?

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nth tems

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How would you solve x^ + y^ = 25 y - 3x = 13

x2 + 2xy = 5 x + y = 3

Hi, I need a bit of help on how to solve the following. y = -4x + 4 y = x2 -6x - 8 I have tried to substitute equation one into equation two ( RHS must be equal) to give -4x + 4 = x2 -6x -8 I have then added all like terms to give but im then stuck and I think i have got the method wrong. Bit confused due to x2 and y = as I thought i did understand how to solve these so im now annoyed. x2 -10x -4 Any help on how to solve this would be helpful. Thanks

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3x-2=-29 6+1/2x=4 X/10+4=5 5x+3+x=75

4(8-x)=40

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3x+2 -------- + 3 = 4 thank you ,:) x-1 the ---- is meant to be divided or over (:

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x + y = 8 x - y = 2

equation: 23-4x=-9 ——- 2

23-4x=-18 is the next step then -4x=18-23 -4x=-18-23 -4x=-41 can i get a step by step answer on how to conver this into x=41/4

please can i have help im really stuck , thanks.