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alexdok [17]
3 years ago
12

Question 2 options: A. y=5x+5 B. y=15x+5 D. Y=1/5x+5 C. Y=10x+5

Mathematics
2 answers:
Gekata [30.6K]3 years ago
8 0
The answer should be A
expeople1 [14]3 years ago
6 0

Answer:

A)

Step-by-step explanation:

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Can someone please help me with this question??
steposvetlana [31]

Answer:

450

Step-by-step explanation:

3*150 because the 150is your main number or otherwise known as "b" and then 3 is the multiplier so 3 * 150

6 0
2 years ago
Can h help with this?
Setler [38]
The Answer would be p(m)=25+10 because she started with a base pay of 10 dollars and the next lawn she charge 35 meaning it makes the rate of change 25
6 0
2 years ago
6x+32=−2x
velikii [3]
Whatever you do to one side, you have to do to the other.
Try to get all the x values and regular values together.
To do that, just subtract both sides by 6x to cancel it out on one side and subtract it from the other side:

6x-6x+32=-2x-6x
=
32=-8x
now divide both sides by -8 to get the x by itself

32/-8=-8/-8x
= 32/-8=x
-4=x
so your answer is: x= -4
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
2 years ago
Kal decides to open a home-based cupcake business. His cost is an initial value of $10 in operating expenses plus $0.25 to make
pshichka [43]

Answer:

4

Step-by-step explanation:

4 0
2 years ago
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