All categories

2 years ago

Linear equations(5x+9)-(7x-1)=

Mathematics

1 answer:

Anuta_ua [19.1K]2 years ago

7 0

-2x+10

You might be interested in

For their twentieth wedding anniversary, Richard and Sylvia had a party for several of their best friends. Things were going alo

kompoz [17]

Answer:

Richard is older

Step-by-step explanation:

We can set up an inequality for both statements.

Let r equal Richard's age and s equal Sylvia's age.

"I am older than my wife."

Since Richard is speaking, the inequality would look like this:

r > s

This means Richard is older than Sylvia.

"I am younger than my husband."

Since Sylvia is speaking, the inequality would look like this:

s < r

This means that Sylvia is younger than Richard.

We can flip one inequality to "see" them from the same perspective.

Let's use s < r

To make it so that we can see the relationship from Richard's perspective, flip the entire inequality.

s < r

r > s

The inequality from the first quote is identical to this one!

Therefore, Richard is older than Sylvia.

8 0

2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

When Nicole works out she spends 10 minutes doing cardio and 15 minutes doing resistance training Isaacs work out consists of 15

Zielflug [23.3K]

Answer:issac

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

What are the independent and dependent variables in the relationship?

Sindrei [870]

w is the width and it is the independent variable. You can pick any positive whole number you want (within reason of course; you can't go to infinity or go beyond some set boundary). Whatever you picked for w, the expression 2w-5 will be dependent on it. So the length is dependent on the width.

For instance, if the width is w = 10 feet, then 2w-5 = 2*10-5 = 20-5 = 15 feet is the length. The choice of 10 feet for the width directly affects the length being 15 feet.

5 0

2 years ago

There were 60 seventh grade students who signed up for soccer tryouts last year. This year, 48 seventh graders signed up for try

Lubov Fominskaja [6]

Answer:

20%

Step-by-step explanation:

(60 - 48) / 60 = 0.2 (20%)

3 0

3 years ago