State the domain of the rational function. f(x) = seventeen divided by quantity five minus x

Mathematics

1 answer:

alexandr1967 [171]4 years ago

4 0

If we are going to write the equation in its mathematical form, we will see that it becomes,

f(x) = 17/(x – 5)

The domain of the function is the number of x that would allow us to solve the equation and get a real value for f(x). In this equation, x can take any real numbers so long as it is not 5. This is because 5 – 5 in the denominator will lead to 0 which in turn makes the equation indefinite.

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Find the particular solution of the differential equation? /=5^3+9^2, when =1, =8

kipiarov [429]

Answer:

$\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}$

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Step-by-step explanation:

*Note:

Ignore the Integration Constant C on the left hand side of the differential equation when integrating.

Step 1: Define

$\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}$

t = 1

s = 8

Step 2: Integrate

[Derivative] Rewrite [Leibniz's Notation]: $\displaystyle ds = (5t^3 + \frac{9}{t^2})dt$
[Equality Property] Integrate both sides: $\displaystyle \int {} \, ds = \int {(5t^3 + \frac{9}{t^2})} \, dt$
[Left Integral] Reverse Power Rule: $\displaystyle s = \int {(5t^3 + \frac{9}{t^2})} \, dt$
[Right Integral] Rewrite [Integration Property - Addition]: $\displaystyle s = \int {5t^3} \, dt + \int {\frac{9}{t^2}} \, dt$
[Right Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt$
[Right Integrals] Rewrite [Exponential Rule - Rewrite]: $\displaystyle s = 5\int {t^3} \, dt + 9\int {t^{-2}} \, dt$
[Right Integrals] Reverse Power Rule: $\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{t^{-1}}{-1}) + C$
[Right Integrals] Rewrite [Exponential Rule - Rewrite]: $\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{1}{t}) + C$
Multiply: $\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} + C$