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igomit [66]
3 years ago
7

I need help!!

Mathematics
1 answer:
Natasha2012 [34]3 years ago
7 0

Answer:

The exact solution is

x = \frac{Ln(60)}{7*Ln(5)}

And the approximation to three decimal places is:

x = 0.363

Step-by-step explanation:

Here we have the equation:

5^{7*x} = 60\\

Now we can remember a property of the natural logarithm function:

Ln(a^n) = n*Ln(a)

Now we can apply the Ln( ) function to both sides of that equation to get:

Ln(5^{7*x}) = Ln(60)

Then we get:

7*x*Ln(5) = Ln(60)

Solving that for x we get:

x = \frac{Ln(60)}{7*Ln(5)}  = 0.363

So the exact solution is:

x = \frac{Ln(60)}{7*Ln(5)}

And the approximation to three decimal places is:

x = 0.363

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Let f(x) = x2 - 81. Find f-1(x).
dusya [7]
This is equivalent to finding a function g in which g(x^2-81)=x.

We simply reverse the actions of the initial function by adding 81 back and taking the square root.  Therefore, f^{-1}(x)=\sqrt{x+81}.
7 0
3 years ago
What two numbers multiply to 10 and add to -29 ?
Degger [83]
Xy=10
x+y=29

subtract x from both sides for second equation

y=-x-29
subsitute in second equaiton
x(-x-29)=10
-x^2-29x=10
add x^2+29x to both sides
x^2+29x+10=0
quadratic formula
if you have
ax^2+bx+c=0,
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}

so
x^2+29x+10=0
a=1
b=29
c=10
x=\frac{-29+/- \sqrt{(-29)^{2}-4(1)(10)} }{2(1)}
x= \frac{-29+/- \sqrt{841-40} }{2}
x= \frac{-29+/- \sqrt{801} }{2}
x= \frac{-29+/- 3\sqrt{89} }{2}

x= \frac{-29- 3\sqrt{89} }{2} or \frac{-29+ 3\sqrt{89} }{2}
those are the 2 numbers
aprox=-28.65 and -0.349


5 0
3 years ago
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Help???? ill take help from anyone
kozerog [31]

Answer:-1/3

Step-by-step explanation: I really don’t know

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2 years ago
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A bag of marbpes contains 6 blue marbles, 2 yellow marbles, 4 red marbles, and 1 green marbpe. What is the probability of reachi
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Answer:

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Step-by-step explanation:

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Setler79 [48]

Answer: inconsistent

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Inconsistent systems have two lines that are two independent equations that don’t overlap. Basically, if the lines are parallel, the system will always be inconsistent.

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