All categories

2 years ago

The function g(x) is defined as g(x)=-2x+3x the value of g(-3)is

Mathematics

2 answers:

Helga [31]2 years ago

5 0

Answer:

-3

Step-by-step explanation:

g(-3) = -2x+3x

-2(-3) +3(-3)

6+-9 = -3

give brainliest please

hope this helps :)

amid [387]2 years ago

4 0

Answer:

just commenting so you can give the other guy brainliest !

Step-by-step explanation:

You might be interested in

Ana pago $350 por 6 pizzas ¿cuánto le costó cada una?

atroni [7]

Usando proporciones y regla de três, tiene-se que cada pizza le costó $58.33.

Para encontrar el precio de cada pizza, usa-se proporciones, aplicando una regla de três.
A medida que aumenta la cantidad, también lo hace el precio, por lo que las medidas son directamente proporcionales y se aplica la multiplicación cruzada.
$350 por 6 pizzas. ¿cuánto por 1?

La regla de três es:

$350 - 6 pizzas

x - 1 pizzas

Aplicando multiplicación cruzada:

$6x = 350$

$x = \frac{350}{6}$

$x = 58.33$

Un problema similar, que tambíem involucra proporciones, es dado en brainly.com/question/24617308

8 0

2 years ago

An integral equation is an equation that contains an unknown function y(x) and an integral that involves y(x). Solve the given i

adelina 88 [10]

Answer:

$y(x)=e^{\frac{-x^{2}}{2}+2 }+1$

Step-by-step explanation:

We have that:

$y(x)=2+\int\limits^x_2 {[t-ty(t)]} \, dt$ (Equation 1)

To resolve this integral equation, we need to use the second Fundamental Theorem of Calculus, which says:

$\frac{d}{dx} [\int\limits^x_a {f(t)} \, dt]=f(x)$

So, we need to differentiate both sides of equation 1 with respect to x:

$\frac{dy}{dx} =\frac{d}{dx} [2+\int\limits^x_2 {[t-ty(t)]} \, dt]$

$\frac{dy}{dx} =\frac{d(2)}{dx}+\frac{d}{dx} [\int\limits^x_2 {[t-ty(t)]} \, dt]$

We know that the derivate for a constant value is zero. And,

$\frac{dy}{dx}=\frac{d}{dx} [\int\limits^x_2 {t} \, dt]-\frac{d}{dx}[\int\limits^x_2 {ty(t)} \, dt]$ (Equation 2)

Using the second Fundamental Theorem of Calculus we know that:

$\frac{d}{dx} [\int\limits^x_2 {t} \, dt]=x\\ \frac{d}{dx} [\int\limits^x_2 {ty(t)} \, dt]=xy(x)$

So, we need to replace those equations in equation 2, and we obtain:

$\frac{dy}{dx} =x-xy(x)$ (Equation 3)

Now, we are going to resolve the equation 3 as a normal equation. So, we need to joint the same variables. I mean, variable y on one side and variable x on other side, as follows:

$\frac{dy}{dx} =x(1-y)\\\frac{dy}{(1-y)} =xdx\\\frac{dy}{y-1} =-xdx$

And, we integrate each side of the equation to obtain:

$ln|y-1|=-\frac{x^{2} }{2} +C$ (Equation 4)

Now, we need to find the value of the constant C. And we know that we can find one point of the equation, replacing x=2 in equation 1, because the integral becomes zero, so:

$y(2)=2+\int\limits^2_2 {t-ty(t)} \, dt=2$