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3 years ago

HELP ASAP!!! EASY POINTS!!!!

Mathematics

2 answers:

Marina CMI [18]3 years ago

4 0

(y-y1)=m(x-x1)

(y-2)=3(x--1)

y=3x--3+2

y=3x+5

3x-y+5=0

notka56 [123]3 years ago

3 0

Answer:

$y-2=3(x+1)$

Step-by-step explanation:

The point-slope formula is $y -y_{1} =m(x -x_{1})$ where we substitute a point (x,y) for $(x_{1},y_{1})$ .

We have m=3 and (-1, 2). We input m and $x_{1} =-1\\y_{1}=2$ .

$y-2=3(x-(-1))\\y-2=3(x+1)$

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4 years ago

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The solid surface area is equal to the solid volume. Find the value of x? The measurements of the rectangular prism is 9 inches

Fudgin [204]

Answer:

$x=7.2$

Step-by-step explanation:

Let x represent height of the prism.

We know that volume of rectangular is equal to product of its height, width and length.

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We have been given that the surface area of rectangular prism is equal to the volume of the prism. So we can set an equation as:

$lwh=2(wl+lh+hw)$

We are also told that the measurements of the rectangular prism is 9 inches long and 4 inches wide. Upon substituting these values in above equation, we will get:

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$36x=2(36+13x)$

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2 years ago

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

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In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

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3 years ago