is a monomial (since it is all connected by multiplication) and since it is in the degree of 2, that makes it a quadratic.
-2 is a constant (since it doesn't have variables) and it is a monomial (since it is all connected by multiplication)
3x-9 is a binomial (since it has two terms, not connected by multiplication) and since it has x in the power of 1, that makes it a linear equation.
Finally,
-6x + 9 is a trinomial (since it has 3 terms connected by either addition or subtraction) and since the degree of the polynomial is 2, that makes it a quadratic.
You would do 6.4/1.6 because big should come first or the number will change
Answer:
D an = 9 + 7(n-1)
Step-by-step explanation:
In the question, Max has 9 logs in the first load. But, after the seventh load he has a total of 51 logs. So the n would be your logs in the stack, which is 51. And since there is a total of 51 logs after the seventh load. It would be 9 plus 7 in parenthesis of 51 minus one, which is 50. So 50 times 7 comes down to 350 plus nine which comes down to a total of 359.
We are asked to find unknown or the missing number to complete the polynomial given in the problem which is x² + ?x -49. First, let us equate the number to be equal to zero such as it would become x² + ?x - 49 = 0. Next, we need to find the factors such that it would produce a difference of squares and these two factors are a = +7 and b = -7. Hence, the complete solution is shown below:
(x + 7) (x-7) = 0
perform distribution and multiplication of terms such as shown below:
x² + 7x - 7x - 49 = 0
Combine the same term such as we can either add or subtract +7x to -7x and the result will be equal to 0x.
x² + 0x - 49 = 0
Therefore, the missing number is 0. The answer is 0 which will result to x² +0x - 49.
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Which is true about the slope of the function y=4x+3?
The slope of the function y=4x+3 is 4; in equations like this one, the slope is simply the number before x.
The rise is 4, and the run is 1.
Remember, slope is also defined as "Rise/Run"
[Note:- You haven't listed any options here, could you kindly add them in the comments?]
<h3>Good luck.</h3>
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