AT&T= 2 years Verizon= 7 years, I hope this helps
Let the price of the tickets be 'x' and the number of tickets bought is 'y'
When the y inversely proportioned to x, the equation is
y = k/x where k is a constant
We work out the value of k by using the information y = 6 and x = 5
6 = k/5
k = 30
Hence, the maximum number of tickets when the price is $3 is
y = 30/3
y = 10 tickets
Answer:
Distributive Property
Step-by-step explanation:
I think it's because the bracket was expanded.
........
4. 14
5. 7/21
........
Rather than trying to guess and check, we can actually construct a counterexample to the statement.
So, what is an irrational number? The prefix "ir" means not, so we can say that an irrational number is something that's not a rational number, right? Since we know a rational number is a ratio between two integers, we can conclude an irrational number is a number that's not a ratio of two integers. So, an easy way to show that not all square roots are irrational would be to square a rational number then take the square root of it. Let's use three halves for our example:
![\sqrt{(\frac{3}{2})^2}=\\\sqrt{\frac{9}{4}}=\\\frac{3}{2}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28%5Cfrac%7B3%7D%7B2%7D%29%5E2%7D%3D%5C%5C%5Csqrt%7B%5Cfrac%7B9%7D%7B4%7D%7D%3D%5C%5C%5Cfrac%7B3%7D%7B2%7D%20)
So clearly 9/4 is a counterexample to the statement. We can also say something stronger: All squared rational numbers are not irrational number when rooted. How would we prove this? Well, let
be a rational number. That would mean,
, would be a/b squared. Taking the square root of it yields:
![\sqrt{\frac{a^2}{b^2}}}=\\ \frac{\sqrt{a^2}}{\sqrt{b^2}}=\\ \frac{a}{b}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%5Cfrac%7Ba%5E2%7D%7Bb%5E2%7D%7D%7D%3D%5C%5C%20%5Cfrac%7B%5Csqrt%7Ba%5E2%7D%7D%7B%5Csqrt%7Bb%5E2%7D%7D%3D%5C%5C%20%5Cfrac%7Ba%7D%7Bb%7D%20)
So our stronger statement is proven, and we know that the original claim is decisively false.