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2 years ago

10

Math help uwu,, please nd thank <3

1 answer:

miskamm [114]2 years ago

3 0

Answer:

Step-by-step explanation:

The equation of a circle is:

(x - h)² + (y - k)² = r²

where (h, k) is the center and r is the radius.

So here, the center is (-2, 4), and the radius is 5.

Here's a graph: desmos.com/calculator/5zqdaam1w2

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7.) There are about 9,987,000 people that live in Michigan. How do you write

aliya0001 [1]

D.) 9,987,000 becomes 9.987 x 10^6

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3 years ago

I need help plz ASAP

netineya [11]

Answer: x=8

Step-by-step explanation:

the sum of the interior angles of a quadrilateral is 360 (number of sides-2 * 180)

so

$14x-7+11x-2+93+76= 360\\25x + 160=360\\25x= 200\\x=8$

4 0

3 years ago

Read 2 more answers

Find the area of the circle. Round to the nearest tenth. Use 3.14 or 22/7 for pie.

maks197457 [2]

Answer:

98.5 cm2

Step-by-step explanation:

6 0

3 years ago

Help!! It's question 14

Leona [35]

Answer:

200/120 = 5/3

5/3 x 3 = 5 oz

<u>you can also do it this way:</u>

3/120 = x/200

x = 600/120 = 5

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