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ddd [48]
3 years ago
8

What is the exponential form of the logarithmic equation? 5 = logo.9 0.59049

Mathematics
2 answers:
OLEGan [10]3 years ago
6 0

Answer:

0.9^5=0.59049

Step-by-step explanation:

By definition, we have:

log_ab=c\implies a^c=b

Therefore, given log_{0.9}(0.59049)=5, assign:

  • a=0.9
  • b=0.59049
  • c=5

Rewrite:

\boxed{0.9^5=0.59049}

galina1969 [7]3 years ago
5 0

^ means to the power of or exponent

logb(x) = y is b^y = x

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Find an equivalent fractions for each given fraction 4/10
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The answer would be 2/5

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10. divided by 2 = 5
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Help with 30 please. thanks.​
Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

\displaystyle  y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}

And we want to show that:

\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}

Instead of differentiating directly, we can first square both sides:

\displaystyle y^2 = 3e^{2x} -4x + 1

We can find the first derivative through implicit differentiation:

\displaystyle 2y \frac{dy}{dx}  = 6e^{2x} -4

Hence:

\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}

And we can find the second derivative by using the quotient rule:

\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}

Substitute:

\displaystyle y\left(\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\right) + \left(\frac{3e^{2x}-2}{y}\right)^2 =6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}- \left(3e^{2x} -2\right)^2}{y^2} + \frac{\left(3e^{2x}-2\right)^2}{y^2}= 6e^{2x}

Combine fractions:

\displaystyle \frac{\left(6y^2e^{2x}-\left(3e^{2x} - 2\right)^2\right) +\left(\left(3e^{2x}-2\right)^2\right)}{y^2} = 6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}}{y^2} = 6e^{2x}

Simplify:

6e^{2x} \stackrel{\checkmark}{=} 6e^{2x}

Q.E.D.

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How do you write 7231÷24
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a_{n} = 5n

Step-by-step explanation:

There is a common difference d between consecutive terms

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a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 5 and d = 5 , then

a_{n} = 5 + 5(n - 1) = 5 + 5n - 5 = 5n

4 0
2 years ago
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