Can someone please answer my recent question

Solve the system of equations x+y=-8 and -9x-6y=60 by combining the equations

TiliK225 [7]

Answer:

x = - 4 y = - 4

Step-by-step explanation:

x+y= - 8

-9x-6y=60

First, solve for x in the first equation:

x+y = - 8 Subtract y from both sides

x + y - y = -8 - y y cancels on the left

x = - 8 - y

Now plug in what you found for x into the 2nd equation and solve for y.

- 9x - 6y = 60

-9(- 8 - y) - 6y = 60 Multiply out

72 + 9y - 6y = 60

72 + 3y = 60 Subtract 72 from both sides

72 - 72 + 3y = 60 - 72 72 cancels on the left

3y = - 12 Divie both sides by 3

3y/3 = -12/3 3 cancels on the left because 3/3 = 1

y = -4

Now plug your answer for y back into the first equation to get x.

x + y = -8

x + (-4) = - 8 Add 4 to each side

x - 4 + 4 = - 8 + 4 4 cancels on the left

x = -4

x = - 4 and y = - 4

8 0

2 years ago

Can someone please send help :(

nlexa [21]

(f o g)(x)

g(x) = x + 2

Since it’s telling you to use g(x), you use (x + 2) as your new x value for function f.

f(x + 2) = 5(x + 2) + 5
= 5x + 10 + 5
= 5x + 15

5 0

2 years ago

1.) A florist sells bouquets for $12.25 each. Amiyah

deff fn [24]

Answer:

Step-by-step explanation:

J

6 0

2 years ago

What does flagfall mean in maths

Alexeev081 [22]

Please explain.
As in taxi fares?

3 0

2 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

6 0

2 years ago