1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Inessa [10]
2 years ago
8

Pls help extra points and mark brainlist 

Mathematics
2 answers:
svlad2 [7]2 years ago
5 0

Answer:

(3,2)

Step-by-step explanation:

...................

prohojiy [21]2 years ago
5 0
(3,2) brainiest? Hope this helps
You might be interested in
HELPPP i need some help solving this out please explain
Eva8 [605]
2x + 2x + x = 180°
5x = 180°
x = 36° (C)
8 0
3 years ago
Read 2 more answers
Find the measure of angle B in isosceles triangle ABC.
jenyasd209 [6]
I think its B)55 
i hope it helps 
3 0
3 years ago
Read 2 more answers
Having trouble with number 5 can someone please help me with it quickly?
Luda [366]

Answer:

I think the answer is 90 degrees

Step-by-step explanation:


3 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
At the movie theatre, child admission is $5.90 and adult admission is $9.10. On Friday, 119 tickets were sold for a total sales
mafiozo [28]

Answer:

47550.501

Step-by-step explanation:

I multiplied them all together but i am not sure if i am correct...

3 0
2 years ago
Other questions:
  • Point P(−5.15, −7.05) is a reflection across the x-axis of which of the following points?PLEASE EXPLAIN
    12·2 answers
  • Can you help me with my algebra
    5·1 answer
  • Tell wheather the system has one solution, infinitely many solutions, or no solution.
    9·1 answer
  • Which statement about this figure is true
    14·2 answers
  • Solve the equation 3×-4y=16 for x
    13·2 answers
  • I need to know math questions to study
    9·1 answer
  • It was recently estimated that homes without children outnumber homes with children by about 8 to 7. If there are 3870 homes in
    7·1 answer
  • The two photos are both for the same question please help I dont know how to solve the revered graph
    6·1 answer
  • Find the measure of an angle in standard position for each reference angle.​
    5·1 answer
  • Liz wants to make several batches of biscuits. She has 27 tablespoons of yeast left in the jar and each batch of biscuits takes
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!