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3 years ago

I need help for 2,3,4 and 5

Mathematics

1 answer:

hoa [83]3 years ago

5 0

Answer:

2 is 12 3 is 3 faces 4 is 151 that's all i can tell u at least i helped

Step-by-step explanation:

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Simplify the expression

cestrela7 [59]

It would come out as 1

7 0

3 years ago

What is the domain of the following parabola?

Montano1993 [528]

The domain of any function is the set of the values of x that will make the function correct. If we try to extend indefinitely the parabola with specified attributes above on the plane where it is drawn, the parabola will expand. This allows all the real numbers be the x-values. Thus, the answer is letter "C. All real numbers"

7 0

3 years ago

The function f is defined by f(x)=2.1x−1.7. Use this formula to find the following values of f. f(x+2)

Yuki888 [10]

The value of $f(x+2)=2.1x+2.5$

Step-by-step explanation:

Given: The function $f(x)=2.1x-1.7$

Using this formula, we need to find the value of $f(x+2)$

Hence, we need to substitute $x$ by $x+2$ to find the value of $f(x+2)$

$f(x+2)=2.1(x+2)-1.7$

Multiplying the term $(x+2)$ by $2.1$ , we get,

$f(x+2)=2.1x+4.2-1.7$

Adding the constant term,

$f(x+2)=2.1x+2.5$

Thus, the value of $f(x+2)$ is $f(x+2)=2.1x+2.5$

8 0

3 years ago

i have 6 digits. One of my 3s is worth 300.000. The other is worth 1/10 as much. My 6 is worth 600.The rest of my digits are zer

motikmotik

The way you wrote the problem makes the answer 330,600. I think your decimals were meant to be commas. If I am wrong, please forgive me.

3 0

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 :

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