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2 years ago

Factorize Зpq - 5р – 2p^2q

Mathematics

1 answer:

NeTakaya2 years ago

6 0

Answer: p x (3q - 5 - 2pq)

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Sandy works at a clothing store. She makes $7 per hour plus earns 10% commission on her sales. She worked 80 hours over the last

maks197457 [2]

Answer:

Option C. $798

Step-by-step explanation:

we know that

The total hours worked the last two weeks ( 80 hours) multiplied by $7 per hour plus the 10% commission on her sales ($2,384) is equal to the amount of money that Sandy earned for those two weeks

Remember that

$10\%=10/100=0.10$

$80(7)+0.10(2,384)$

$\$560+\$238.4=\$798.40$

7 0

2 years ago

What is 4.236 divided by 6

neonofarm [45]

0.706 is the answer to your problem

3 0

2 years ago

Jaelynn has 30 times as many stickers as her brother. Her brother has 8 stickers. How many stickers does Jaelynn have?

Dennis_Churaev [7]

30 x 8 = 240 so Jaelynn has 240 stickers :) hope I helped!

6 0

3 years ago

Read 2 more answers

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: