Kate is going to purchase a coat for $38, pants for $45, and 2 pairs of shoes for $34 each. if she has $180 to spend how much mo
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Answer:
To prove that 3·4ⁿ + 51 is divisible by 3 and 9, we have;
3·4ⁿ is divisible by 3 and 51 is divisible by 3
Where we have;
= 3·4ⁿ + 51
= 3·4ⁿ⁺¹ + 51
-
= 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ
-
= 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ
∴ -
is divisible by 9
Given that we have for S₀ = 3×4⁰ + 51 = 63 = 9×7
∴ S₀ is divisible by 9
Since -
is divisible by 9, we have;
-
=
-
is divisible by 9
Therefore is divisible by 9 and
is divisible by 9 for all positive integers n
Step-by-step explanation:
We will conclude that:
- The domain of the exponential function is equal to the range of the logarithmic function.
- The domain of the logarithmic function is equal to the range of the exponential function.
<h3>Comparing the domains and ranges.</h3>
Let's study the two functions.
The exponential function is given by:
f(x) = A*e^x
You can input any value of x in that function, so the domain is the set of all real numbers. And the value of x can't change the sign of the function, so, for example, if A is positive, the range will be:
y > 0.
For the logarithmic function we have:
g(x) = A*ln(x).
As you may know, only positive values can be used as arguments for the logarithmic function, while we know that:
So the range of the logarithmic function is the set of all real numbers.
<h3>So what we can conclude?</h3>- The domain of the exponential function is equal to the range of the logarithmic function.
- The domain of the logarithmic function is equal to the range of the exponential function.
If you want to learn more about domains and ranges, you can read: