1. t+7=5 step by step
1 answer:
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Answer:
14.52% are unicyclists
Step-by-step explanation:
First, we can use proportions to find the number of unicyclists at the convention. Since we know that the ratio of unicyclists to aerial artists is 9:11, and there are 88 aerial artists, we can set up the following equation:
If we cross multiply, we get that 11x = (88)(9). After we divide through by 11 to isolate x, we get that x = (8)(9) = 72
Second, we have to figure out the number of mimes at the convention to figure out the total number of people there. We know that the ratio of unicyclists to mimes is 3:14, and the number of unicyclists is 72. So, we can set up the following proportion:
If we cross multiply, we get that 3y = 1008, or y = 336 mimes
The total number of people at the convention is 336 mimes + 72 unicyclists + 88 unicyclists = 496. Now we have to figure out what percent of 496 is 72 (the number of unicyclists). If we let z = the percentage, we can simply set up an equation that says that 72 is z% of 496:
This means that approximately 14.52% of the performers are unicyclists.
Answer:
See below.
Step-by-step explanation:
Let's let our first integer be n.
Then, our second, consecutive integer must be (n+1).
We want to prove that the sum of the square of two consecutive integers is always odd.
So, let's square our two expressions and add them up:
Square. Use the perfect square trinomial pattern. So:
Combine like terms:
Notice that the first term 2n² has a 2 in front. Anything multiplied by 2 is even. So, regardless of what integer n is, 2n² is <em>always</em> even.
Our second term 2n also has a 2 in front. So, whatever n is, 2n is also <em>always</em> even.
So far, 2n² and 2n are both even. Therefore, the sum of 2n² and 2n is <em>also </em>even.
Lastly, we have the constant term 1. It doesn't matter what n is, we know that (2n² + 2n) is definitely and is always even. So, if we add 1 to this, we will get an odd number.
Q.E.D.