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3 years ago

What is the domain of the step function f(x) = [2x] – 1?

Mathematics

2 answers:

omeli [17]3 years ago

6 0

You have troubles with domains only if you have:

Denominators
Roots with an even degree
Logarithms

In this case, you have neither of these three, so you can compute your function wherever you want. The domain is all real numbers.

Leya [2.2K]3 years ago

4 0

Answer:

d all real numbers......:)

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Which equation could have been used to create this graph?

Mekhanik [1.2K]

B.) y=3x
The slope of a given line is defined by y=mx+b, where b is the y-intercept (the point at which the line crosses the y-axis) and m is the slope (rise/run). We see that the y-intercept is 0 (because the line hits the y-axis at 0) and the slope is 3 (rise=3 and run=1, so 3/1=3).

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3 years ago

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What goes into the square root of 96?

VladimirAG [237]

Hi there, the factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96. So, we want the highest factor of 96. The factor we will use is the number 16. Lets get solving. √96=√16*6=√16*√6=4√6, Therefore, the square root of 96 simplified is 4√6 and the square root of 96 is 9.79795897113

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3 years ago

Read 2 more answers

(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 :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

2 years ago

Write an expression that represents the area of a circle with radius 3x^3. (For any circle with radius r, a=π r^2 = π .r .r ^.)

aalyn [17]

9514 1404 393

Answer:

B. 9πx^6

Step-by-step explanation:

Putting the given radius into the area formula gives ...

A = πr^2

A = π(3x^3)^2 = π(3^2)(x^(3·2))

A = 9πx^6

6 0

2 years ago

A pizza place offers 3 different cheeses and 12 different toppings. In how many ways can a pizza be made with 1 cheese and 3 top

svetlana [45]

You have to choose 1 sort of cheese from 3 ( 3 ways ). Also you have to choose 3 toppings from 12 different toppings:
12 * 11 * 10 = 1,320 ways.
In total: 3 * 1,320 = 3,960
Answer:
In 3,690 ways can pizza be made with 1 cheese and 3 toppings.

3 0

3 years ago