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3 years ago

Find the slope between the following points, show all work. (5, -9) and (-9, -5)

2 answers:

6 0

To find your slope you have to put your points in this formula y2-y1/x2-x1, lets plug in the numbers -5+9/-9-5 now add or subtract your numbers to get 4/-14

Hope this helps

Sloan [31]3 years ago

5 0

The answer to the question is -2/7

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5(y-4)=7(2y+1) I need the answer

8_murik_8 [283]

Answer: $y = -3$

Move all terms containing y to the left, all other terms to the right

Add -14y to each side of the equation

$-20 + 5y + -14y = 7 + 14y + -14y$

Combine like terms: 5y + -14y = -9y

$-20 + -9y = 7 + 14y + -14y$

Combine like terms: 14y + -14y = 0

$-20 + -9y = 7 + 0\\-20 + -9y = 7$

Add '20' to each side of the equation

$-20 + 20 + -9y = 7 + 20$

Combine like terms: -20 + 20 = 0

$0 + -9y = 7 + 20\\-9y = 7 + 20$

Combine like terms: 7 + 20 = 27

$-9y = 27$

Divide each side by -9

$-9y/9=27/-9\\y=-3$

8 0

3 years ago

1/3 x 1/2 PLS PSL eellelele

Darina [25.2K]

1/6
explanation:
1 x 1 = 1
2 x 3 = 6
1/6
Hope it helps!

6 0

3 years ago

Read 2 more answers

[100 POINTS] What is the measure of the arc from A to B that does not pass through C?

Nana76 [90]

Applying the inscribed angle theorem, the measure of arc AB that doesn't go through point C is: 100 degrees.

<h3>What is the Inscribed Angle Theorem?</h3>

Based on the inscribed angle theorem, if ∅ is the inscribed angle measure, the measure of the central angle subtended by the same arc equals 2(∅).

m∠BAC = 40 degrees.

Central angle = 2(40) = 80 degrees [based on the inscribed angle theorem]

Corresponding arc BC = 80 degrees.

Arc AC through point B = 180 degrees [half circle]

Arc AB = 180 - arc BC = 180 - 80 = 100 degrees.

Learn more about the inscribed angle theorem on:

brainly.com/question/3538263

#SPJ1

8 0

1 year ago

Find the balance of the account using the simple interest formula of I=Prt

cestrela7 [59]

I don’t know what the formula stands for but if it was in years instead of months I could totally answer

3 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

2 years ago