Answer:
Charlie has used his phone in a month for at least 1404 minutes
Step-by-step explanation:
In order to solve this problem, we must first determine what will our variable be and what it will represent.
Let's say our variable is x and it will represent the number of minutes Charlie has used his phone.
After we set our variable up, we can set our equation up. The problem states that Charlie will pay a monthly fee of $18 and additional $0.06 per minute of use. The $18 is what is called a fixed cost and the $0.06 is the variable cost, which will depend on our variable x (the number of minutes spent). Taking this into account we can build an inequality that will represent the amount of money spent in a month, which will look like this:
so now we can solve that inequality for x, we can start by subtracting 18 from both sides, so we get.
Next, we can divide both sides of the inequality by 0.06 so we get:
so that's where the answer came from. Charly has used an amount of at least 1404 minutes
Answer:
![1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}](https://tex.z-dn.net/?f=1.%20%5Cquad%5Cdfrac%7B1%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%5C%5C%5C%5C2.%20%5Cquad%5Csqrt%5B7%5D%7Bx%5E5%7D%5C%5C%5C%5C3.%20%5Cquad%5Cdfrac%7B1%7D%7B%5Csqrt%5B5%5D%7By%5E2%7D%7D)
Step-by-step explanation:
The applicable rule is ...
![x^{\frac{m}{n}}=\sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%3D%5Csqrt%5Bn%5D%7Bx%5Em%7D)
It works both ways, going from radicals to frational exponents and vice versa.
The particular power or root involved can be in either the numerator or the denominator. The transformation applies to the portion of the expression that is the power or root.
Answer:
the number is 25
Step-by-step explanation:
If the number is n, then twenty-two more is n + 22
which equals 47, so n+ 22 = 47
n = 47 - 22 = 25
Standard equation of a circle: <em>(x-h)² + (y-k)² = r²</em> where <em>(h, k)</em> is the center and <em>r </em>is the radius. In the case of our equation here, <em>(x-5)² + (y+3)² = 25</em>, we can conclude that our circle has a center at (5, -3) and a radius of 5 units.
We can use the distance formula with the center (5, -3) and our point (2, 3) to see how far away they are...if the distance between them is less than the radius of the circle, it is on the interior. If it's equal, it's on the circle. If it's greater, it's on the exterior.
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