Actually, 40 is divisible by 7. Any number is divisible by 7. However, 7 does not go into forty evenly. Your quotient would have a remainder since 7 cannot go into 40 an even number of times.

7 goes into 40 5 times evenly (7x5=35) 40-35 is five, so 40 divided by 7 is 5 (remainder 5)

~theLocoCoco

6 0

2 years ago

Determine whether the given differential equation is exact. If it is exact, solve it. (If it is not exact, enter NOT.) (tan(x) −

never [62]

Answer:

$f(x,y)=ln secx+cosx siny+C$

Step-by-step explanation:

We are given that DE

$(tanx-sinx siny)dx+cosxcosydy=0$

We have to determine given DE is exact or not.

Compare it with Mdx+Ndy=0

$M=tanx-sinx siny$

$N=cosxcosy$

$\frac{\partial M}{\partial y}=$ $M_y=-sinxcosy$

$\frac{\partial N}{\partial x}=N_x=-sinxcosy$

Therefore, $M_y=N_x$

If DE is exact then $M_y=N_x$

Hence,it is exact.

$M=\frac{\partial f}{\partial x}=tanx-sinxsiny$

Integrate w.r.t x on both sides

$f(x,y)=\int(tanx-sinxsiny)dx$

$f(x,y)=lnsecx+cosxsiny+\phi(y)$ ...(1)

By using the formula

$\int tanxdx=lnsecx+C,\int sinxdx=-cosx+C$

Differentiate partially equation (1) w.r.t y

$\frac{\partial f}{\partial y}=cosxcosy+\phi'(y)$

By using the formula:

$\frac{d(sinx)}{dx}=cosx$

$N=\frac{\partial f}{\partial y}=cosxcosy=cosxcosy+\phi'(y)$

$\phi'(y)=cosxcosy-cosxcosy=0$

$\phi'(y)=0$

Integrate w.r.t y

$\phi(y)=C$

Substitute the value

$f(x,y)=ln secx+cosx siny+C$

7 0

3 years ago

Eighth grade Y.14 Solve equations with the distributive property SRP Solve for u. -9.6u - 6.96 = -3(2.4u – 4.72) – = U = Submit​

Eighth grade Y.14 Solve equations with the distributive property SRP Solve for u. -9.6u - 6.96 = -3(2.4u – 4.72) – = U = Submit