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3 years ago

Solve for x in the literal equation -20=xy +z.

Mathematics

1 answer:

erastovalidia [21]3 years ago

5 0

Answer:

-(20+z)/y = x

Step-by-step explanation:

-20=xy +z

Subtract z from each side

-20 -z = xy+z-z

-20 -z = xy

Divide each side by y

(-20 -z)/y = xy/y

Factor out a negative

-(20+z)/y = x

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A patio was to be laid in a design with one tile in the

Vlada [557]

Answer:

Step-by-step explanation:

The number of tiles in the design is 1 + 2 + 3 + ...

We can model this as an arithmetic series, where the first term is 1 and the common difference is 1. The sum of the first n terms of an arithmetic series is:

S = n/2 (2a₁ + d (n − 1))

Given that a₁ = 1 and d = 1:

S = n/2 (2(1) + n − 1)

S = n/2 (n + 1)

Since S ≤ 60:

n/2 (n + 1) ≤ 60

n (n + 1) ≤ 120

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n ≤ 10

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2 years ago

Given the coordinates for the function below which of the following are coordinates for its inverse

Diano4ka-milaya [45]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

The value of Q is given by the relation Q = 17m. If Q and m are integers with Q greater than 150 and less than 160, then which o

zloy xaker [14]

Answer: Value of m = 9

Step-by-step explanation:

Given that the relationship between Q and m is;

Q = 17m

Make m the subject of formula

M = Q/17

If Q is greater than 150 and less than 160, then, let assume that

Q = 151, then

M = 151/17

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2 years ago

First ten nonzero multiples of 4,5,6,7,10

ArbitrLikvidat [17]

4 times table: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
5 times table: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
6 times table: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
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10 times table: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100

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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

Solve for x in the literal equation -20=xy +z.​

Solve for x in the literal equation -20=xy +z.