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3 years ago

Use slope and y intercept to graph a line. y=2x-5

2 answers:

8 0

On your graph:

-- Mark a dot on the y-axis, at y = -5 .

-- From there, move 1 unit to the right and 2 units up and make a mark. If this
is too tiny for you, then you can move 7 units to the right and 14 units up, or
11 units to the right and 22 units up ... any way you want to do it, as long as
the distance 'up' is double the distance to the right, because the slope is 2 to 1.
Wherever you wind up, mark a dot.

-- Using your pencil and your ruler, draw a straight line between the two dots you have
marked. You may extend it as far as you wish in either or both directions.

Naily [24]3 years ago

6 0

M=2,b=-5 where m is the slope and b is the y-intercept

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Which equation has the same meaning as y=log8 x?<br> 8^x=y<br> x=y^8<br> x=8y<br> x=8^y

a_sh-v [17]

$\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\[-0.35em] ~\dotfill\\\\ y=\log_8(x)\implies 8^y=x$

6 0

3 years ago

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VladimirAG [237]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

Use the method of cylindrical shells to find the volume v generated by rotating the region bounded by the curves about the given

kondor19780726 [428]

Answer:

The volume is $\frac{16\pi}{e}$

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

rectangle

- If each vertical rectangle is revolved about the y-axis, we

obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

of the cylinder times the thickness of the cylinder

- The formula for the volume will be: $V=\int\limits^b_a {2\pi xf(x)} \, dx$

where 2πx · f(x) is the surface area of the cylinder shell and dx is its

thickness

* Lets solve the problem

- To find the volume V generated by rotating the region bounded

by the curves y = 4e^x and y = 4e^-x about the y-axis by use

cylindrical shells

- Consider that the height of the cylinder is y = (4e^x - 4e^-x)

- Consider that the radius of the cylinder is x

- The limits are x = 0 and x = 1

- Lets take 2π and 4 as a common factor out the integration

∴ $V=\int\limits^1_0 {2\pi x(4e^{x}-4e^{-x})} \, dx$

∴ $V=2\pi(4)\int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx$

- To integrate $xe^{x}$ and $xe^{-x}$ we will use

integration by parts methods $\int\ {uv'=uv-\int{v}\,u' }\,$

∵ u = x

∴ u' = du/dx = 1 ⇒ differentiation x with respect to x is 1

∵ v' = dv/dx = e^x

- The integration e^x is e^x ÷ differentiation of x (1)

∴ $v=\int\ {e^{x}}\, dx= e^{x}$

∴ $\int\ {xe^{x}} \, dx=xe^{x}-\int\ e^{x}\, dx=xe^{x}-e^{x}$

- Similar we will integrate xe^-x

∵ u = x

∴ u' = du/dx = 1

∵ v' = dv/dx = e^-x

- The integration e^-x is e^x ÷ differentiation of -x (-1)

∴ $v=\int\ {e^{-x}} \, dx=-e^{-x}$

∴ $\int\ {x}e^{-x}\, dx=-xe^{-x}+\int\ {e^{-x}} \, dx=-xe^{-x}-e^{-x}$

∴ V = $8\pi \int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx=8\pi[xe^{x}-e^{x}+xe^{-x}+e^{-x}]$ from 0 to 1

- Lets substitute x = 1 minus x = 0

∴ $V=8\pi[(1)(e^{1})-(e^{1})+(1)(e^{-1})+(e^{-1})-(0)(e^{0})+(e^{0})-(0)(e^{0})-(e^{0})]$

∴ $V=8\pi[e^{1}-e^{1}+e^{-1}+e^{-1}-0+1-0-1]=8\pi[2e^{-1}]=16\pi e^{-1}$

∵ $e^{-1}=\frac{1}{e}$

∴ $V=\frac{16\pi}{e}$

3 0

4 years ago

In a pet store, the small fishbowl holds up to 23

kvv77 [185]

Answer: the large fishbowl hold up to 46 gallons of water

Step-by-step explanation:

4 0

3 years ago

A 13-foot piece of string is cut into two pieces so that the longer piece is 4 feet longer than twice the shorter piece. Find t

Yuri [45]

The answer is 10 and 3
6/2 = 3
6 + 4 = 10
10 + 3
it said one piece was twice the other plus 4.

4 0

3 years ago