5 1/2 % in simplest form

Diego plays center forward on his soccer team. On average, he scores 1.25 goals per match. The tally of his total goals for the

masya89 [10]

F(x) = 1.25(38 - x)

Finding the inverse of a function:
y = f(x)

y = 1.25(38 - x)

Solve for x:
y = 1.25(38 - x)
y/1.25 = 38-x
x = 38 - y/1.25

Switch x and y.

y = 38 - x/1.25

f^-1(x) = 38 - x/1.25

3 0

3 years ago

Danielle has her finals coming up and she's worried she'll fail her class. Her past 3 test grades were a 72. an 85, and a 68. Sh

natka813 [3]

Answer:

55

Step-by-step explanation:

7 0

3 years ago

If f(x) = x² + 1 for what x value is f(x) = 2?

MaRussiya [10]

Answer:

x = ±1

Step-by-step explanation:

Step 1: Define variables

f(x) = x² + 1

f(x) = 2

Step 2: Substitute

2 = x² + 1

Step 3: Solve for <em>x</em>

0 = x² - 1

0 = (x - 1)(x + 1)

x - 1 = 0

x = 1

x + 1 = 0

x = -1

∴ x = ±1

6 0

3 years ago

Help plz ill give extra points

UkoKoshka [18]

Answer:

79 degrees

Step-by-step explanation:

PQU is a straight line, so $\angle PQU = 180^\circ$ . This is also the sum of the four smaller angles. We can subtract 34+37+30 from 180 to get that $\angle SQT = 79^\circ$

(if right pls give brainliest :) )

5 0

3 years ago

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

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