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1 answer:
It is true that the product of two consecutive even integers are always one less than the square of their average.
<u>Step-by-step explanation</u>:
Let the two consecutive odd integers be 1 and 3.
- The product of 1 and 3 is (1
3)=3
- The average of 1 and 3 is (1+3)/2 =4/2 = 2
- The square of their average is (2)² = 4
∴ The product 3 is one less than the square of their average 4.
Let the two consecutive even integers be 2 and 4.
- The product of 2 and 4 is (2
- The average of 2 and 4 is (2+4)/2 =6/2 = 3
- The square of their average is (3)² = 9
∴ The product 8 is one less than the square of their average 9.
Thus, It is true that the product of two consecutive even integers are always one less than the square of their average.
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Answer:
Step-by-step explanation:
Another complex expression, let's simplify it step by step...
We'll start by re-writing 256 as 4^4
Then we'll extract the 4 from the cubic root. We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.
Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.
For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.
The answer is then:
Use the following identities:
Also because the angle is in quadrant 3, sin must be negative.
Therefore
Subbing in tan = 0.958
Also because the angle is in quadrant 3, sin must be negative.
Therefore
Subbing in tan = 0.958