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Strike441 [17]
3 years ago
7

1. Complete the inequality below to describe the

Mathematics
2 answers:
dezoksy [38]3 years ago
5 0

Domain is the entire span left to right (on the x-axis) that the graph is on. Since the graph goes from x=-4 and ends at x=4, the domain would be from -4 to 4. The circle at -4 is open, so it does not include the point at -4, just everything leading up to it. So, the domain would be

- 4 < x \leqslant 4

The range is similar, it is the entire span that the graph goes up and down (on the y-axis). The graph starts at the bottom at y=-2, and ends at y=5. The bottom point (4,-2) is closed, so the graph includes that point, and the top point (-4,5) is open and doesn't include the point. Therefore, the range would be

- 2 \leqslant x < 5

a_sh-v [17]3 years ago
4 0

Answer:

your answer is

<u><em>A</em></u>

<u><em>B</em></u>

Step-by-step explanation:

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Solve the differential equation. y' + 5xey = 0.
galben [10]

Answer:

The solution is     y = - ln(\frac{5}{2}x^{2} + C)

Step-by-step explanation:

To solve the differential equation, we will find y

From the given equation, y' + 5xey = 0.

That is, y' + 5xe^{y} = 0

This can be written as

\frac{dy}{dx} + 5xe^{y} = 0

Then,

\frac{dy}{dx} = - 5xe^{y}

\frac{dy}{e^{y}}   = - 5x dx

Then, we integrate both sides

\int {\frac{dy}{e^{y}}}  =\int {- 5x dx}

\int {e^{-y}dy }}  =\int {- 5x dx}

Then,

-e^{-y} = -\frac{5}{2}x^{2} + C

e^{-y} = \frac{5}{2}x^{2} + C

Then,

ln(e^{-y}) = ln(\frac{5}{2}x^{2} + C)

Then,

-y = ln(\frac{5}{2}x^{2} + C)

Hence,

y = - ln(\frac{5}{2}x^{2} + C)

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3 years ago
Find the value of Y to nearest tenth... <br> cos 64degrees = y/8
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Step-by-step explanation:

what is the problem ? this is very straight forward with a calculator :

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2 years ago
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