Ask question

Ask question

All categories

3 years ago

7

Sasha runs 8 miles on Thursday, then runs 8 more miles on Friday, she ran 1/2 mile on Saturday. hHOW MANY MILES DID SHE RUN? PLE

ASE HELP IM BEING TIMED

2 answers:

Over [174]3 years ago

3 0

Answer:

16 1/2 miles

Step-by-step explanation:

djyliett [7]3 years ago

3 0

It would be 16 1/2 miles

You might be interested in

What is the logarithmic form of 11² = 121?

Alona [7]

Logbase11(121)=2. The base of the log to the power of y (outside number) = the number in the parenthesis (x).

7 0

3 years ago

Read 2 more answers

PLZZZZ HELPPPPPP NOWWWWWW PLZ PLZ PLZ

telo118 [61]

Answer:

1. T

2. O

3. L

4. C

5.B

6.H

Step-by-step explanation:

8 0

2 years ago

What is the cosine of 45°

Degger [83]

Cosine 45° = √2/2 (Square root of 2 over 2)

4 0

3 years ago

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

!PLS HELP I WILL GIVE BRAINLEST!

AysviL [449]

The answer would be 18 square units

Step-by-step explanation:

When talking about surface area, just add up all the units that is listed in the question. - just a tip ;)

Anyways, Hope this helps!! If it's wrong, feel free to curse me out.. haha...

6 0

3 years ago