Answer:
d = 14/5
Step-by-step explanation:
The point (-4,2) means that;
At x = -4, y = 2
Now general form of a linear equation is;
Ax + By + C = 0
We are given;
4y = 3x + 6
Rearranging to the form of a linear equation gives;
3x - 4y + 6 = 0
Thus, A = 3, B = -4 and C = 6
Thus, at point (-4,2), distance between them is;
d = (3(-4) - 4(2) + 6)/√(3² + (-4)²)
d = -14/5
We will take the absolute value.
Thus; d = 14/5
Hey there!
For this problem, I would assume that you must make your own chart of values. You can do this by simply plugging in numbers like 1, 2, 3, etc... Since your graph won't be linear, you might want to do some negative numbers, as well.
Also because your graph isn't linear, you might want to figure out or determine on your own just how far your curve will go. You can then figure out how many times you need to plug in an additional numbers for your function to then graph.
I've attached a digital graph of this function, perhaps you can find it useful.
Hope this helped you out! :-)
Answer:
2
Step-by-step explanation:
Answer: A, B and
Explanation:
Angle 3, 4, and 2 are the exterior angles because it locates outside
Multiplying both sides by
gives

so that substituting
and hence
gives the linear ODE,

Now multiply both sides by
to get

so that the left side condenses into the derivative of a product.
![\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Bx%5E3v%5D%3D3x%5E2)
Integrate both sides, then solve for
, then for
:




![\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}](https://tex.z-dn.net/?f=%5Cboxed%7By%3D%5Csqrt%5B3%5D%7B1%2B%5Cdfrac%20C%7Bx%5E3%7D%7D%7D)