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3 years ago

HURRY PLEASEEEEE!!!

Mathematics

1 answer:

Sever21 [200]3 years ago

4 0

Answer:

5.5 or 5 or 6

Step-by-step explanation:

anne=5+7=12(blankets per day)

bella=2+8=10 (blankets per day)

We need to use the 12 times table and 10 times table and see when they meet up

12, 24, 36, 48, 60

10, 20, 30, 40, 50, 60

it will take 5.5 days or 5 or 6

You might be interested in

(4x+3y^2)dx+2xy*dy=0 by using integrating fector

Nataly_w [17]

$\underbrace{(4x+3y^2)}_M\,\mathrm dx+\underbrace{2xy}_N\,\mathrm dy=0$

The ODE is exact if $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$ .

$M_y=6y$
$N_x=2y$

This is not the case, so look for an integrating factor $\mu(x)$ such that

$\dfrac\partial{\partial y}\mu M=\dfrac\partial{\partial x}\mu N$

Since $\mu$ is a function of $x$ only, you have

$\mu M_y=\mu'N+\mu N_x\implies\dfrac{\mu'}\mu=\dfrac{M_y-N_x}N\implies\mu=\exp\left(\displaystyle\int\frac{M_y-N_x}N\,\mathrm dx\right)$

So, the integrating factor is

$\mu=\exp\left(\displaystyle\int\frac{6y-2y}{2xy}\,\mathrm dx\right)=\exp\left(2\int\frac{\mathrm dx}x\right)=x^2$

Now the ODE can be modified as

$\underbrace{(4x^3+3x^2y^2)}_{M^*}\,\mathrm dx+\underbrace{2x^3y}_{N^*}\,\mathrm dy=0$

Check for exactness:

${M^*}_y=6x^2y$
${N^*}_x=6x^2y$

so the modified ODE is indeed exact.

Now, you're looking for a solution of the form $\Psi(x,y)=C$ , since differentiating via the chain rule yields

$\dfrac{\mathrm d}{\mathrm dx}\Psi(x,y)=\Psi_x+\Psi_y\dfrac{\mathrm dy}{\mathrm dx}=0$

Matching up components, you would have

$\Psi_x=M^*=4x^3+3x^2y^2$
$\displaystyle\int\Psi_x\,\mathrm dx=\int(4x^3+3x^2y^2)\,\mathrm dx$
$\Psi=x^4+x^3y^2+f(y)$

Differentiate this with respect to $y$ to get

$\Psi_y=2x^3y+f'(y)=2x^3y=N^*$
$f'(y)=0\implies f(y)=C_1$

So the solution here is

$\Psi(x,y)=x^4+x^3y^2+C_1=C\implies x^4+x^3y^2=C$

Just for a final check, take the derivative to get back the original ODE:

$\dfrac{\mathrm d}{\mathrm dx}[x^4+x^3y^2]=\dfrac{\mathrm d}{\mathrm dx}C$
$4x^3+3x^2y^2+2x^3y\dfrac{\mathrm dy}{\mathrm dx}=0$
$4x+3y^2+2xy\dfrac{\mathrm dy}{\mathrm dx}=0$
$(4x+3y^2)\,\mathrm dx+2xy\,\mathrm dy=0$

so the solution is correct.

7 0

4 years ago

What is the solution of the system of equations

White raven [17]

Answer:

It the 3rd one it's correct cuz i worked it out

Can i have a brainliest please!!:)

Step-by-step explanation:

solve your system by substitution.

c+3d=8;c=4d−6

Rewrite equations:

c=4d−6;c+3d=8

Step: Solve c=4d−6for c:

c=4d−6

Step: Substitute4d−6forcinc+3d=8:

c+3d=8

4d−6+3d=8

7d−6=8(Simplify both sides of the equation)

7d−6+6=8+6(Add 6 to both sides)

7d=14

(Divide both sides by 7)

d=2

Step: Substitute2fordinc=4d−6:

c=4d−6

c=(4)(2)−6

c=2(Simplify both sides of the equation)

Answer:

c=2 and d=2

6 0

3 years ago

PLEASE GIVE ME AN ANSWER QUICKLY! Hiroshi spends 30 minutes on history homework , 60 minutes on English homework, and x minutes

natka813 [3]

Answer:

Step-by-step explanation:

He spends:
30 mins on History
60 mins on English
x mins on math
so the total time he spends is $(TotalTime)minutes=(30+60+x)minutes$
one fourth of this is his math time, since x is math time, then we have $\frac{1}{4}(TotalTime)minutes=(x)minutes$
But we know what totaltime in minutes is so $\frac{1}{4}(30+60+x)minutes=(x)minutes$
or simply: $\frac{1}{4}(30+60+x)=x$