Not of Bernoulli type, but still linear.
There's no need to find an integrating factor, since the left hand side already represents a derivative:
![\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}+2xy](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B%281%2Bx%5E2%29y%5D%3D%281%2Bx%5E2%29%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B2xy)
So, you have
![\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=4x^2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B%281%2Bx%5E2%29y%5D%3D4x%5E2)
and integrating both sides with respect to
yields
Answer:
Not sure if this is correct but this is what I found.
area of sector=74.84
radius = 8
angle of sector=?
74.84=1/2×8²×angle of sector
74.84=32 × angle of sector
74.84/32= angle of sector
2.33875=angle of sector
Step-by-step explanation:
Sorry if it is incorrect.
First, we can convert both of them to improper fractions.
We do that by multiplying the denominator to the whole number, adding it to the numerator, and keeping the denominator.
2 5/3 - 2 3/2
So we have:
11/3 - 7/2
Convert both of them to denominators of 6:
22/6 - 21/6
Subtract the numerators and keep the denominators:
1/6
For part a,
y < x
y > -2x
Point B: (3,1) plugin x=3, y=1 y < x 1 < 3
which is true, so point B satisfies the first inequalityYou can graph the inequality y > 2x+5 and see which schools are in the shaded region
Answer:
7.1
Step-by-step explanation:
Since it is a square you can use the Pythagorean theorem (a²+b²=c²) by filling in what you know. a²=5 and b²=5 so you have 5²+5²=c². This simplfies to 25+25=c². Now you simplify again to get 50=c². Finally you have to square root everything because the equation can't end in c², it has to be the most simplified it can be. This equals ≈7.1
