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Alexxx [7]
3 years ago
14

Which equation is a point slope form equation for line AB ?

Mathematics
1 answer:
Jobisdone [24]3 years ago
3 0

For this case we have that by definition, the equation of a line of the point-slope form is given by:

y-y_ {0} = m (x-x_ {0})

Where:

m: It's the slope

(x_ {0}, y_ {0}):It is a point through which the line passes

To find the slope, we need two points through which the line passes, observing the image we have:

(x_ {1}, y_ {1}): (1,6)\\(x_ {2}, y_ {2}): (5, -2)\\m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-2-6} {5-1} = \frac {-8} {4} = -2

Thus, the equation is of the form:

y-y_ {0} = - 2 (x-x_ {0})

We choose a point:

(x_{0}, y_ {0}) :( 5, -2)

Finally, the equation is:

y - (- 2) = - 2 (x-5)\\y + 2 = -2 (x-5)

Answer:

y + 2 = -2 (x-5)

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(b) Jo travels 4 kilometres in 1/2 hour = 9 km/hour.

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Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

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\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
3 years ago
There are 50 deer in a particular
olchik [2.2K]

answered

There are 50 deer in a particular forest. The population is increasing at a rate of 15% per year. Which exponential growth function represents

the number of deer y in that forest after x months? Round to the nearest thousandth.

1

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1

dribeiro

Ace

697 answers

561.4K people helped

Answer:

The expression that represents the number of deer in the forest is

y(x) = 50*(1.013)^x

Step-by-step explanation:

Assuming that the number of deer is "y" and the number of months is "x", then after the first month the number of deer is:

y(1) = 50*(1+ 0.15/12) = 50*(1.0125) = 50.625

y(2) = y(1)*(1.0125) = y(0)*(1.0125)² =51.258

y(3) = y(2)*(1.0125) = y(0)*(1.0125)³ = 51.898

This keeps going as the time goes on, so we can model this growth with the equation:

y(x) = 50*(1 - 0.15/12)^(x)

y(x) = 50*(1.013)^x

8 0
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