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2 years ago

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5. 4i(-2 - 4i)

1 answer:

Wewaii [24]2 years ago

4 0

Answer:

$4i( - 2 - 4i) = - 8i + 16 \\ \\ 5i.i.( - 2i) = 10i \\ \\ (- 5 - i)(1 + i) = -5-5i-i+1\\=-4-6i$

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If a measured distance of 5 inches on a map respresents an actual distance of 3 miles, what is the fractional scale of the map?

OverLord2011 [107]

5 inches: 3 miles
1 inch: 0.6 miles

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3 years ago

HELP I AM STUCK AND ITS DUE SOON

Mila [183]

Answer:

I'm not for sure about this one but if I did my math right it should be 68 sence it would be 48 if you didnt add 20

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3 years ago

Read 2 more answers

Write an algebraic expression for the verbal expression<br> the quotient for 45 and r

Nimfa-mama [501]

Answer:

45/r

Step-by-step explanation:

You're just doing 45 divided by r.

4 0

2 years ago

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

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

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

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

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

A falling object accerlates from -10.0m/s to -30.0m/s how much time does that take

Leya [2.2K]

Answer:

2.04 seconds

Step-by-step explanation:

Falling objects near the surface of the earth have an acceleration of -9.81 m/s².

Acceleration is the change in velocity over change in time:

a = (v − v₀) / t

-9.81 = (-30.0 − (-10.0)) / t

-9.81 = -20.0 / t

t = 2.04

It takes 2.04 seconds.

5 0

3 years ago