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3 years ago

-8= -2(z+7) How do you sovle this equation

Mathematics

2 answers:

Vitek1552 [10]3 years ago

7 0

-8=-2z-14
+2z-8=-14
2z=-14+8
2z=-6
2z\2z=-6/2z
z=3

Anna71 [15]3 years ago

5 0

$-8= -2(z+7)$ Equation

$\dfrac{-8}{-2} =z+7

 
$ $Divide \ both \ sides \ by \ -2$

<span> $\dfrac{8}{2} =z+7$
</span>
$4=z+7

 
$

$4-7=z$ $Subtract \ 7 \ from \ both \ sides$

$z=-3$ $Solution$

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3 years ago

Jun used 56 cm of ribbon to go around the edge of a rectangular lid. The lid was 13 cm long. What is the width of the lid?

maksim [4K]

Answer:

The width of the rectangular lid is 15 cm.

Step-by-step explanation:1. Let's review the information given to us to answer the question correctly:

Length of the ribbon used by Jun = 56 cmLength of the rectangular lid = 13 cm

2. What is the width of the lid? The ribbon used by Jun is equivalent to the perimeter of the rectangular lid, therefore:Perimeter = 2 * Length + 2 * WidthReplacing with the values we have:56 = 2 * 13 + 2 * Width56 = 26 + 2Width2Width = 56 - 262Width = 30Width = 30/2Width = 15

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2 years ago

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Consider the equation

earnstyle [38]

a) true

b) false

c) true

Step-by-step explanation:

<h3>let's determine the first statement</h3><h3>to determine x-intercept </h3><h3>substitute y=0</h3>

so,

8x-2y=24

8x-2.0=24

8x=24

x=3

therefore

the first statement is <u>true</u>

let's determine the second statement

<h3>to determine y-intercept </h3><h3>substitute x=0</h3>

so,

8x-2y=24

8.0-2y=24

-2y=24

y=-12

therefore

the second statement is <u>False</u>

to determine the third statements

<h3>we need to turn the given equation into this form</h3><h2>y=mx+b</h2><h3>let's solve:</h3>

8x-2y=24

-2y=-8x+24

y=4x-12

therefore,

the third statement is also <u>true</u>

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2 years ago

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vazorg [7]

Answer:

t=0.8

Step-by-step explanation:

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3 years ago

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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

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

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

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

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

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3 years ago