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Leno4ka [110]
3 years ago
12

Evaluate xe^(x^2+y^2+z^2) where e is the portion of the unit ball x^2+y^2+z^2 < 1 that lies in the firat octant

Mathematics
1 answer:
AnnyKZ [126]3 years ago
8 0
Given

\int\int\int_E xe^{(x^2+y^2+z^2)}

where E <span>is the portion of the unit ball x^2+y^2+z^2\leq1 that lies in the first octant.

This can be evaluated as follows:

\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}} xe^{(x^2+y^2+z^2)}dzdydx=0.392699</span>
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Do you compare a negative decimal to a positive decimal?
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Answer:

Positive: Decimals are a part of a whole just like fractions are a part of a whole. Therefore, a positive decimal is always greater than a negative decimal.

Negative: When you have two negative decimals, the one closer to zero is always greater. The farther a negative decimal is from zero, the smaller its value.

All of it put together (same text as above):

Decimals are a part of a whole just like fractions are a part of a whole. Therefore, a positive decimal is ALWAYS greater than a negative decimal. When you have two negative decimals, the one closer to zero is always greater. The farther a negative decimal is from zero, the smaller its value.

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3 years ago
HELP PLEASE!!!!!!!!!!!!
Bogdan [553]

Answer:

See the explanation.

Step-by-step explanation:

We are given the function f(x) = x² + 2x - 5

Zeros :

If f(x) = 0 i.e. x² + 2x - 5 = 0

The left hand side can not be factorized. Hence, use Sridhar Acharya formula and  

x= \frac{-2+\sqrt{2^{2}-4\times(-5)\times1 } }{2} and  

x= \frac{-2-\sqrt{2^{2}-4\times(-5)\times1 } }{2}

⇒ x = -3.45 and 1.45

Y- intercept :

Putting x = 0, we get, f(x) = - 5, Hence, y-intercept is -5.

Maximum point :

Not defined

Minimum point:  

The equation can be expressed as (x + 1)² = (y + 5)

This is an equation of parabola having the vertex at (-1,-5) and axis parallel to + y-axis

Therefore, the minimum point is (-1,-5)

Domain :  

x can be any real number

Range:  

f(x) ≥ - 6

Interval of increase:

Since this is a parabola having the vertex at (-1,-5) and axis parallel to + y-axis.

Therefore, interval of increase is +∞ > x > -1

Interval of decrease:

-∞ < x < -1

End behavior :  

f(x) = x^{2} +2x-5 =x^{2}  (1+\frac{2}{x} -\frac{5}{x^{2} } )

So, as x tends to +∞ , then f(x) tends to +∞

And as x tends to -∞, then f(x) tends to +∞. (Answer)

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