A couch went on a 16% off sale. The sale price is given as $1008. What was the original price of the couch?

Mathematics

1 answer:

scoray [572]2 years ago

4 0

Answer:

1200 dollars

Step-by-step explanation:

x-0.16x=1008

.84x=1008

x=1200 dollars

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Solving radical equations

My name is Ann [436]

Answer:

top: no solution
bottom: x = 7

Step-by-step explanation:

I find a graphing calculator very helpful to solving radical equations. It helps avoid extraneous solutions.

top

√(-14x +2) = x -3

-14x +2 = x² -6x +9 . . . . . square both sides

x² +8x +7 = 0 . . . . . . . . . .subtract the left side

Descartes' rule of signs tells you this has no positive real solutions. (There are no sign changes among the coefficients.) The value of the radical will be non-negative, so we must have x-3≥0, or x≥3. Since there are no positive real solutions to the quadratic, there are no values of x≥3 that will satisfy the equation. NO SOLUTION

bottom

4√(x -3) = 8 . . . . . given

√(x -3) = 2 . . . . . . divide by 4

x -3 = 4 . . . . . . . . . square both sides

x = 7 . . . . . . . . . . . add 3

3 0

4 years ago

Your uncle is 2 years older than 3 Times your age you are x years old x=2 write an expression to describe your uncles age you ar

kirza4 [7]

Because you are 12 you put down

12

x can be anything so if you write

12x

You can later substitute it for 3. 12(3)

Your uncle is 2 years more

12x+2 is your equation
or
12(3) + 2

3 0

4 years ago

Read 2 more answers

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$