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marshall27 [118]

2 years ago

10

Find the measure of AD⏜ in ⊙F ? LESSON Measuring Angles and Arcs Explain your reason

1 answer:

Svetllana [295]2 years ago

6 0

Answer:

153

Step-by-step explanation:

Angle AFE is 63 and Angle DFE is a 90 degree angle add them both up and you get 153

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VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV this is middle school math

Yuri [45]

Answer:

first method

<u>2</u><u>2</u>=<u>5</u>

7. x

we cross multiply to get

<u>2</u><u>2</u><u>x</u>= <u>5</u><u>×</u><u>7</u>

22. 22.

x= <u>3</u><u>5</u>

22

=1.59

second method

<u>2</u><u>2</u>=<u>5</u>

7. x

we multiply the denominators to get 7x

then we multiply each term by 7x

7x×<u>2</u><u>2</u> = <u>5</u>×7x

7. x

here the 7 and 7 will cancel out and the x and x will cancel out to get

<u>2</u><u>2</u><u>x</u>= <u>3</u><u>5</u>

22. 22

= 1.59

5 0

3 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

Sergio comma Eduardo comma Maria comma Simone comma Ian comma Dawn comma Larry comma and Tyrone have all been invited to a dinne

ANTONII [103]

Answer:

a.

Step-by-step explanation:

5 0

3 years ago

The graph shown is the solution set for which of the following inequalities?

elena55 [62]

The inequality is y < l x+3 l

8 0

3 years ago

Read 2 more answers

Determine if -10,-12,or -16is the solution of 2t-2=26

Dafna11 [192]

2t-2=26
first you would add 2 to both sides
that equals 2t=28
then you would divide both sides by two
that would give you
t= 14

3 0

3 years ago