All categories

2 years ago

How can the zero product be applied to real life

Mathematics

1 answer:

ddd [48]2 years ago

7 0

Answer:

The zero product property states that if a⋅b=0 then either a or b equal zero. This basic property helps us solve equations like (x+2)(x-5)=0.

The Zero Product Property simply states that if ab=0 , then either a=0 or b=0 (or both). A product of factors is zero if and only if one or more of the factors is zero.

Step-by-step explanation:

hope it helps

You might be interested in

Robert has a number cube, a coin, and a (fair) spinner numbered 1 to 6.

Elan Coil [88]

I think B because I think B

5 0

3 years ago

Determine the values of the constants r and s such that i(x, y) = x rys is an integrating factor for the given differential equa

garri49 [273]

$\underbrace{y(7xy^2+6)}_{M(x,y)}\,\mathrm dx+\underbrace{x(xy^2-1)}_{N(x,y)}\,\mathrm dy=0$

For the ODE to be exact, we require that $M_y=N_x$ , which we'll verify is not the case here.

$M_y=21xy^2+6$
$N_x=2xy^2-1$

So we distribute an integrating factor $i(x,y)$ across both sides of the ODE to get

$iM\,\mathrm dx+iN\,\mathrm dy=0$

Now for the ODE to be exact, we require $(iM)_y=(iN)_x$ , which in turn means

$i_yM+iM_y=i_xN+iN_x\implies i(M_y-N_x)=i_xN-i_yM$

Suppose $i(x,y)=x^ry^s$ . Then substituting everything into the PDE above, we have

$x^ry^s(19xy^2+7)=rx^{r-1}y^s(x^2y^2-x)-sx^ry^{s-1}(7xy^3+6y)$
$19x^{r+1}y^{s+2}+7x^ry^s=rx^{r+1}y^{s+2}-rx^ry^s-7sx^{r+1}y^{s+2}-6sx^ry^s$
$19x^{r+1}y^{s+2}+7x^ry^s=(r-7s)x^{r+1}y^{s+2}-(r+6s)x^ry^s$
$\implies\begin{cases}r-7s=19\\r+6s=-7\end{cases}\implies r=5,s=-2$

so that our integrating factor is $i(x,y)=x^5y^{-2}$ . Our ODE is now

$(7x^6y+6x^5y^{-1})\,\mathrm dx+(x^7-x^6y^{-2})\,\mathrm dy=0$

Renaming $M(x,y)$ and $N(x,y)$ to our current coefficients, we end up with partial derivatives

$M_y=7x^6-6x^5y^{-2}$
$N_x=7x^6-6x^5y^{-2}$

as desired, so our new ODE is indeed exact.

Next, we're looking for a solution of the form $\Psi(x,y)=C$ . By the chain rule, we have

$\Psi_x=7x^6y+6x^5y^{-1}\implies\Psi=x^7y+x^6y^{-1}+f(y)$

Differentiating with respect to $y$ yields

$\Psi_y=x^7-x^6y^{-2}=x^7-x^6y^{-2}+\dfrac{\mathrm df}{\mathrm dy}$
$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

Thus the solution to the ODE is

$\Psi(x,y)=x^7y+x^6y^{-1}=C$

4 0

3 years ago

8. | 11 | 14 | 17 | 20<br> What’s the function

Murrr4er [49]

Answer:

What do you exactly need to do?

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

Please answer this correctly without making mistakes

Gekata [30.6K]

Answer:

Step-by-step explanation:

The distance is given by

d = sqrt( (x2-x1) ^2 + ( y2-y1) ^2)

Since the x values are the same

d = sqrt( ( y2-y1) ^2)

d = sqrt( (( -6 - 10)^2))

d = sqrt((-16) ^2)

d = 16

We can also just find the number of squares between the points since the x value is the same.

10 - -6

10+6

4 0

3 years ago

Read 2 more answers

Find the area of the composite figure.

AysviL [449]

Answer:

The area will be 834.38

Step-by-step explanation:

hope it helps..

5 0

3 years ago