You have a math test. you plan to study 5/6 hour tonight and 3/4 hour tomorrow.

A piece of plywood was cut so it's length with 8 ft by 4 ft what is the area of the wood

Elena L [17]

The area of an object is found by multiplying the length by the width, so to solve your problem, multiply 8*4 to get 32.

8 0

3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

A baker has 6 small bags of flour. Each bag weighs 1 pound. She divides each bag into thirds. How many 1/3 - pound bags of flour

lana [24]

Answer:

18 smaller bags

Step-by-step explanation:

Each one pound bag becomes 3 one-third pound bags.

So 6 one pound bags will become 6×3=18 one-third pound bags.

5 0

4 years ago

In the figure, AB is parallel to CD, XY is the perpendicular bisector of AB, and E is the midpoint of XY. Prove that △AEB ≅ △DEC

Makovka662 [10]

Answer:

From top to bottom:

A, J, E, B, I, C, D, G, F, H

See below for more clarification.

Step-by-step explanation:

We are given that AB is parallel to CD, XY is the perpendicular bisector of AB, and E is the midpoint of XY. And we want to prove that ΔAEB ≅ ΔDEC.

Statements:

1) XY is perpendicular to AB.

Definition of perpendicular bisector.

2) XY ⊥ CD.

In a plane, if a transveral is perpendicular to one of the two parallel lines, then it is perpendicular to the other.

3) m∠AXE = 90°, m∠DYE = 90°.

Definition of perpendicular lines.

4) ∠AXE ≅ ∠DYE.

Right angles are congruent.

5) XE ≅ YE

Definition of a midpoint.

6) ∠A ≅ ∠D.

Alternate Interior Angles Theorem

7) ΔAEX ≅ ΔDEY

AAS Triangle Congruence*

(*∠A ≅ ∠D, ∠AXE ≅ ∠DYE, and XE ≅ YE)

8) AE ≅ DE

Corresponding parts of congruent triangles are congruent (CPCTC).

9) ∠AEB ≅ ∠DEC

Vertical Angles Theorem

10) ΔAEB ≅ ΔDEC

ASA Triangle Congruence**

(**∠A ≅ ∠D, AE ≅ DE, and ∠AEB ≅ ∠DEC)

3 0

3 years ago

How to identify the domain and range of a function?

sergiy2304 [10]

Domain are the values of x, range are the values of y

3 0

3 years ago