A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function. -The quadratic coefficients is -The linear coefficients is -The constant term is

So my course shows me three differential equations:
$\dot{x}+x^2=t$
$\dot{x}=(t^2+1)(x-1)$
$\dot{x}+x=t^2$
The first one is not a linear ordinary differential equation (ODE) apparently, the other two are.
Unfortunately, they don't show a clear way how to find out if an ODE is linear or not. So how we can find out if an ODE is linear?
For the second one, I thought I bring it into standard form somehow:
$\dot{x}=(t^2+1)(x-1)=x{t}^2+x-t^2-1=x(t^2+1)-t^2-1$
If we say we let $p(t)=t^2+1$ and $q(t)=1+t^2$, then we could say:
$\dot{x}=xp(t)-q(t)=...$
And so on, to simplify until we reach standard form of a linear ODE (or not).
Is that the way to go? Or is there some other way to check if a ODE is linear?

-6n - 20 = -2n + 4(1 - 3n)

The scale on a picture is 0.5mm : 40mm. the height of a bottle in the picture is 4.5mm. What is the actual height of the bottle in cm?

To evaluate: The value of the expression |-10|+(-62).

Given:
${\displaystyle a{x}^2+bx+c=0}$, with a,b and c being whole numbers, and this equation having two solutions. The solutions of such an equation are sometimes rational, but mostly not. How do you prove that the sum of the solutions ${\displaystyle x_1+x_2}$ and the product ${\displaystyle x_1\cdot x_2}$ are always rational?

Solve the given system of quadratic equations
${\displaystyle x^2+xy+xz=2}$
${\displaystyle y^2+yz+xy=3}$
${\displaystyle z^2+zx+yz=4}$

To translate the given phrase into algebraic expression
A number subtracted from ${\displaystyle -\frac{8}{9}}$