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3 years ago

Find the values of p and q such that 4x^2+12x=4(x+p)^2-q

Mathematics

1 answer:

-Dominant- [34]3 years ago

7 0

Answer:

p = 1.5 and q = 9

Step-by-step explanation:

Expand the right side then compare the coefficients of like terms on both sides, that is

4(x + p)² - q ← expand (x + p)² using FOIL

= 4(x² + 2px + p²) - q ← distribute parenthesis

= 4x² + 8px + 4p² - q

Comparing coefficients of like terms on both sides

8p = 12 ( coefficients of x- terms ) ← divide both sides by 8

p = 1.5

4p² - q = 0 ( constant terms ), that is

4(1.5)² - q = 0

9 - q = 0 ( subtract 9 from both sides )

- q = - 9 ( multiply both sides by - 1 )

q = 9

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Tell me, what is 24 x 0.25?

Neko [114]

Answer:

it is 6.

Step-by-step explanation:

4 0

2 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

Solve 2x - 8 + 3x > 5x - 4 also if you can plz explain

WITCHER [35]

Answer:

No solution

Step-by-step explanation:

2x - 8 + 3x > 5x - 4

2x + 3x - 8 > 5x - 4

5x - 8 > 5x - 4

-5x +8 > -5x +8

----------------------

0x > 4

This is no solution, as there is no 'x' value.

(0x just means there isn't anything left, I put that in order to show there wasn't anything.)

7 0

3 years ago

Find the length of the unknown side round your answer to the nearest whole number 18 and 4

EastWind [94]

Answer:

17.5

Step-by-step explanation:

90% sure

8 0

2 years ago

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frosja888 [35]

Answer:

106.25

Step-by-step explanation:

y = 85(1.25) = 106.25

I'm not sure how to show the work, i just put it into a calculator

6 0

2 years ago