HELP PLEASE Find the value of y. Round to the nearest tenth. у 42° 85° 31

Which is larger 1 or 2

My name is Ann [436]

1 is larger than 2 .

3 0

3 years ago

(4)

Jet001 [13]

Answer:

Step-by-step explanation:

Let the number of cockles be c and that of winkles be w.

From the first relation,

w = 2c ..........(I)

From the second relation, 6 winkles are accidentally split.

This in equation form means the following:

c/w- 6 = 3/5

By cross multiplication,

5c = 3(w - 6)...........(ii)

We know w = 2c , let’s substitute this into the second equation.

5c = 3(2c - 6)

5c = 6c - 18

Collecting like times, c = 18

And w = 2c = 2(18) = 36

Hence, there were initially 18 cockles and 36 winkles

7 0

3 years ago

Can someone please help on 11&12?

Svet_ta [14]

# 12 should be the last one

8 0

3 years ago

In a class, 7 students walk to school, two fifths ride the bus, and 32% are dropped off by their parents. How many students ride

34kurt

Answer:

About 40% I think-

Step-by-step explanation:

Basically, you change 2/5 into a percentage and you should get 40 percent.

8 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