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2 years ago

A small cube has the volume shown. Its side length is 1.5 in less than a second larger cube. What is

Mathematics

1 answer:

eduard2 years ago

6 0

The volume of the larger cube is 227 inches cube.

<h3 /><h3 /><h3>Volume of a cube is describe below:</h3>

v = L³

where

L = length

The small cube has the following volume:

v = 95 in³

Let

The side length of the small cube = x

Therefore,

95 = x³

x = ∛95

x = 4.56290264

x ≈ 4.6

Its side length is 1.5 inches less than a second larger cube. Therefore,

length of larger length cube = (x + 1.5 )inches = 4.6 + 1.5 = 6.1 inches

Therefore, the volume of the larger cube can be found as follows:

Volume = 6.1³

volume = 226.981

volume ≈ 227 inches³

learn more on cubes here:brainly.com/question/2671825?referrer=searchResults

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Answer:

Choice A) $F(x) = 3\sqrt{x + 1}$ .

Step-by-step explanation:

What are the changes that would bring $G(x)$ to $F(x)$ ?

Translate $G(x)$ to the left by $1$ unit, and
Stretch $G(x)$ vertically (by a factor greater than $1$ .)

$G(x) = \sqrt{x}$ . The choices of $F(x)$ listed here are related to $G(x)$ :

Choice A) $F(x) = 3\;G(x+1)$ ;
Choice B) $F(x) = 3\;G(x-1)$ ;
Choice C) $F(x) = -3\;G(x+1)$ ;
Choice D) $F(x) = -3\;G(x-1)$ .

The expression in the braces (for example $x$ as in $G(x)$ ) is the independent variable.

To shift a function on a cartesian plane to the left by $a$ units, add $a$ to its independent variable. Think about how $(x-a)$ , which is to the left of $x$ , will yield the same function value.

Conversely, to shift a function on a cartesian plane to the right by $a$ units, subtract $a$ from its independent variable.

For example, $G(x+1)$ is $1$ unit to the left of $G(x)$ . Conversely, $G(x-1)$ is $1$ unit to the right of $G(x)$ . The new function is to the left of $G(x)$ . Meaning that $F(x)$ should should add $1$ to (rather than subtract $1$ from) the independent variable of $G(x)$ . That rules out choice B) and D).

Multiplying a function by a number that is greater than one will stretch its graph vertically.
Multiplying a function by a number that is between zero and one will compress its graph vertically.
Multiplying a function by a number that is between $-1$ and zero will flip its graph about the $x$ -axis. Doing so will also compress the graph vertically.
Multiplying a function by a number that is less than $-1$ will flip its graph about the $x$ -axis. Doing so will also stretch the graph vertically.

The graph of $G(x)$ is stretched vertically. However, similarly to the graph of this graph $G(x)$ , the graph of $F(x)$ increases as $x$ increases. In other words, the graph of $G(x)$ isn't flipped about the $x$ -axis. $G(x)$ should have been multiplied by a number that is greater than one. That rules out choice C) and D).

Overall, only choice A) meets the requirements.

Since the plot in the question also came with a couple of gridlines, see if the points $(x, y)$ 's that are on the graph of $F(x)$ fit into the expression $y = F(x) = 3\sqrt{x - 1}$ .

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3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

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