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3 years ago

15

Two streets intersect at a 30 degree angle. At the intersection, there are four crosswalks formed that are the same length. what

type of quadrilateral is formed by the crosswalks?

1 answer:

ELEN [110]3 years ago

3 0

Answer:

rhombus

Step-by-step explanation:

A quadrilateral with equal-length sides is a <em>rhombus</em>.

_____

A square is a special case of a rhombus.

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(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 <em>µ(x, y)</em> 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 <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

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

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 <em>F(x, y)</em> = <em>C</em>. 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 <em>x</em> :

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

Differentiate both sides of this with respect to <em>y</em> :

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

Then the general solution to the DE is

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

5 0

2 years ago

Which of the expressions below has a 6 in the hundredths place of the product? Select all that apply.

Sliva [168]

Answer:

A

Step-by-step explanation:

7 0

3 years ago

The five members of the Traynor family each buy train tickets. During the train ride, each family member buys a boxed lunch for

larisa [96]

Let's assume

the price of each train ticket is x

we are given

total number of family members =5

so, total cost of ticket = total number of family members*the price of each ticket

so, total cost of ticket =5*x

so, total cost of ticket =5x

now, we have

each family member buys a boxed lunch for $6.50

so, total cost of lunch box = total number of family members*price of each lunch box

so, total cost of lunch box = 5*6.50

total cost of trip = total cost of ticket+total cost of lunch box

we are given

total cost of trip is $248.50

now, we can plug values

and we get

$248.50=5x+5*6.50$

now, we can solve for x

$248.50=5x+32.5$

$248.50-32.5=5x+32.5-32.5$

$216=5x$

$\frac{5x}{5} =\frac{216}{5}$

$x=43.2$

so,

the price of each train ticket is $43.2..........Answer

8 0

3 years ago

Read 2 more answers

What does subsequent mean

Olenka [21]

Subsequent means following (in time, order or place f.e.)

6 0

2 years ago

What is the slope of a line that is perpendicular to the line whose equation is 0.52 - 5y = 9?

otez555 [7]

The slope is 0.1
To find a perpendicular slope, find the negative reciprocal.
The reciprocal of 0.1 is 10
The negative of 10 is -10
Final answer:-10

7 0

2 years ago