N+N=30 what is the number for N

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

y= x2 (2x+1)1/2

find the equation of the line below arrange your answer in the form y=mx+b where b is constant (-2,2) (2,2)

x4/3 - 102/3 +21=0

find the equation of the tangent line to (e^xy)+(x^2)-(y^2)=5 through the point (2,0)

9x+5=7x+11

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

3x⁴-2x³-5x²-4/x²-4

how do i figure out x2+7x+10=0

please can you give me a step by step guide

2(3x+8)=-23+9

how do i solve 2(3b - 7) = 4 by expanding brackets?

how do quadratic equations work?

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

f(x)=-3(x+5)(x-11)

how to factor 6y^2=29y+9

How to solve equation ln(x)=ln(x+2)-1

How to solve (3x-4)=(4x-3)2

i've got to............ write down 2 expressions for the perimeter of a square one with brackets and another without brackets of t+3

If george is 378 of the age of his brother, and in 4 years he will be 1/2 of the age of his brother. What is the age of the brother?

Solve the simultaneous equations: x2 +2 y2 = 8 and x + 2y =8

I need help with this please help?

2x - 3y = 21 5x + 2y = 5

A gas bill consists of a fixed charge (£F) and charge (g pence) for each unit used. Mrs Ali used 350 units and paid 30 pounds. mr hill used 450 units and paid £35. find the fixed charge and the charge per unit.

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

I really need help with topic. I don't get it at ALL. this is one of the questions used in class:- y=2x-1. y=x2 (Both x and y equal to 1 I just don't understand how or why) please explain

Sorry here is the problem 8x-(3x+10)=34

how do i solve x for (2/x+3)=9/4=5 and none is in parenthesis.

(X-2) over 3 + 1 = x over 7

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

4(8-x)=40

Solve for x. 50 - 32 % =x

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

solve the simultaneous equations, anyone help please?!

please can i have help im really stuck , thanks.

stimulateneous eqation 2x + 4y =1 3x-5y=7

Use the formula A=p(1+r)n (is the exponent) to find out how much [/latex]500 would be worth in 20 years if

its an algebra question

what is 3(x-4)

can you please answer this question in 10 mins please!!

what is the answer to y=x+9 ?

what is the growth factor and the initial value for the Koch snowflake?

what is the product of 7 and 122

What is x and y

Solving Quadratic Simultaneous Equations

Try to figure out how to solve this equation: x2 + y2 = 29 y - x = 3

How to Find the Answer

Lets begin by rewriting our two equations, x²+y²=29…..iy-x=3……….ii Next we take our equation ii and equate y in terms of x as follows, y-x=3y=x+3… Now we substitute y term in equation i:. x²+y²=29 where there’s the value of y we replace it with x+3 x²+(x+3) (x+3) =29 we rewrite the equation in quadratic formatx²+x²+3x+3x+9=292x²+6x+9=29 2x²+6x+9-29=02x²+6x-20=0……now we have formed our quadratic equation. To solve the equation we may use the factorization method or completing square method. So in the case here,  I prefer to use the completing square method because our equation doesn’t have perfect squares. Which is [- b±(b²_4ab)¹/²]/2a Next, we substitute the coefficients of x in our equation above, where a=2, b=6, and c=-20[-(6)±(6²-{4×2×-20})¹/²]/2×2{-6±(36+160)¹/²}/4 {-6±(196)¹/²}/4 where the root of 196=14(-6±14)/4x=(-6+14)/4=2x=(-6-14)/4=-5 Now we have the possible values of x 2&-5, then we substitute these values on our y equation y=x+3y=2+3y=5y=-5+3y=-2 Therefore our values for both x and y will be x=2 or – 5y=5 or – 2 This is because our simultaneous equation had a quadratic equation.

(X-2) over 3 + 1 = x over 7