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3 years ago

NO LINKS!!!!!! Please please help me!!! ASAP

Mathematics

1 answer:

nydimaria [60]3 years ago

5 0

Answer:

PLEASE TAKE YOUR TIME AND THINK THROUGH EVERY PROBLEM. USE THE C.U.B.E.S STRATEGY WHEN NEEDED.

Step-by-step explanation:

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A shopping bag contains six orange juice boxes and four grape juice boxes. Without

juin [17]

Answer:

6/25 i think

Step-by-step explanation:

5 0

3 years ago

The circumference of a circle is 15π ft. What is the area, in square feet? Express your answer in terms of π.

DerKrebs [107]

Answer:

Area of the circle is 56.25π square feet.

Step-by-step explanation:

Given that:

The circumference of a circle is 15π ft.

To Find:

What is the area, in square feet?

We know that:

Circumference of a circle = 2πR
Area of a circle = πR²

Where,

Radius is denoted as R.

Finding the radius of the circle:

Circumference of a circle = 15π

⟶ 2πR = 15π

⟶ 2πR = 2π × 7.5

Cancelling 2π.

⟶ R = 7.5

∴ Radius of the circle = 7.5 ft.

Finding the area of the circle:

⟶ Area = πR²

⟶ Area = π(7.5)²

⟶ Area = π × 7.5 × 7.5

⟶ Area = π × 56.25

⟶ Area = 56.25π

∴ Area of the circle = 56.25π ft.²

7 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

Suppose that x and y vary inversely, and x = 12 when y = 8 . Write the function that models the inverse variation.

Alja [10]

Answer:

$y=\frac{96}{x}$

Step-by-step explanation:

So when x and y, vary inversely, it means that the product of x and y will remain constant such that: $xy=c$ . This means that the function can be defined as $y=\frac{c}{x}$ by dividing both sides by x. In this case x=12, and y=8, this means the constant is equal to 12 * 8 which is 96. So we have the equation: $xy = 96$ , which can be defined as $y=\frac{96}{x}$

7 0

1 year ago

PLZ HURRY IT'S URGENT!

const2013 [10]

I = prt

To solve for r divide both sides by p and t.

r = I /pt

6 0

3 years ago