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Vilka [71]
3 years ago
8

Candy is on sell for $0.75 each. You have a coupon for $.25 off your total purchase. Write a function rule for the cost of n pie

ces of candy
Mathematics
2 answers:
weeeeeb [17]3 years ago
8 0

Answer:

.75 -.25=n

Step-by-step explanation:

.75 -.25=n

enyata [817]3 years ago
6 0

t(n)=.75n-.25

Term number x .75 subtract .25



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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x
Serggg [28]

I assume C has counterclockwise orientation when viewed from above.

By Stokes' theorem,

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

so we first compute the curl:

\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k

\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k

Then parameterize S by

\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k

where the z-component is obtained from

1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v

with 0\le u\le\dfrac\pi2 and 0\le v\le\dfrac\pi2.

Take the normal vector to S to be

\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k

Then the line integral is equal in value to the surface integral,

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}

6 0
3 years ago
What is the slope of this line?
Anvisha [2.4K]
(-4,5),(0,6)

slope = rise/run
tip... you have to crawl (run) before you run (rise) .

you can use the slope formula
\frac{5 - 6}{ - 4 - 0}  =   \frac{ - 1}{ - 4}  =  \frac{1}{4}
1/4
5 0
3 years ago
Look at the box-and-whisker plot.
Marizza181 [45]

i would say B

when it says "the measure", it's refering to where the second quartile is beginning

3 0
3 years ago
Kim solved the equation below by graphing a system of equations. log base 2 (3x-1) = log base 4 (x+8)
kompoz [17]

Answer:

solution is

x=1.353,y=1.613

Step-by-step explanation:

We are given equation as

log_2(3x-1)=log_4(x+8)

Firstly, we will find equations

First equation is

y=log_2(3x-1)

Second equation is

y=log_4(x+8)

now, we can draw graph

and then we can find intersection point

we can see that

intersection point is (1.353,1.613)

so, solution is

x=1.353,y=1.613

8 0
3 years ago
Read 2 more answers
I need help on this question
schepotkina [342]

Answer:

  0 4 9 1

             _____________

1 9      9 3 2 9

 − 0      

   9 3    

 − 7 6    

   1 7 2  

 − 1 7 1  

       1 9

     − 1 9

         0

Step-by-step explanation:

Sorry of this isn't clear

6 0
3 years ago
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