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3 years ago

What is the domain of the following function?

Mathematics

1 answer:

ki77a [65]3 years ago

7 0

Answer:

{-1, 0, 1, 2, 3, 4, 5}

Step-by-step explanation:

The domain is all the possible X values so on this graph it would be all the x values from -1 - -5

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What is the area of a semicircle with a diameter of 14

Neporo4naja [7]

Answer:

76.9690 square units (rounded off to four decimal values)

Step-by-step explanation:

The diameter of the semicircle is 14 units

The radius of the semicircle is thus 14 ÷ 2 = 7 units

Area of a semicircle is given by: $\frac{1}{2}$ × π × r²

In our case, the area is = $\frac{1}{2}$ × π × 7² = 76.9690200129 square units

Or 76.9690 square units (rounded off to four decimal values)

8 0

3 years ago

Read 2 more answers

In Union County, Florida, the 1990 census listed the population at 10,252. The 2000 census listed the population as 13,442. What

lora16 [44]

Answer:

31.1% increase

Step-by-step explanation:

% increase= 100 X (final-initial)/initial

5 0

3 years ago

the histogram shows the result of survey asking students how many windows are in their homes how mnay studengs have less than 10

serious [3.7K]

Answer:

4 students

Step-by-step explanation:

Although I am not sure if I am completely correct. I’ll try my best.

We know:

The numbers 0-12 on the frequency table is the number of windows.

The numbers on the bottom is the amount of students.

What we can see:

As we can see..

the 0-4 section in the table shows that only 0-4 students have 9 windows.
The 5-9 sections shows that only 5-9 students have 10 windows.
The 10-14 sections shows that 10-14 students have 0 windows.
The 15-19 sections shows that 15-19 students have 4 windows.
The 20-24 sections shows that 20-24 students have 2 windows.

Calculations:

Since the number of students that have the amount of windows less than 10 is 4 (0-4 section) and 0 (10-14 section)

Add.

4+0

= 4

Therefore, 4 students have less than 10 windows.

4 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 :

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