Answer:
6c^2-cd-d^2
Step-by-step explanation:
How you get the answer is you use the process called FOIL. The F is front, so you multiply the 2c and the 3c to get 6c^2. The O is the outer, so you multiply the 2c and the d(because they are on the outer sides of the parenthesis) to get 2cd. The I is the inner, so you multiply the -d and the 3c(because they are on the inner sides of the parenthesis) to get -3cd. The L is the last, so you multiply d and -d to get -d^2. The 2cd and the -3cd are the same so you have to minus them to get -cd. So the answer is:
6c^2-cd-d^2.
Hello!
log₃(x) + log₃(x - 6) = log₃(7) <=>
<=> log₃(x * (x - 6)) = log₃(7) <=>
<=> log₃(x² - 6x) = log₃(7) <=>
<=> x² - 6x = 7 <=>
<=> x² - 6x - 7 = 0 <=>
<=> x² + x - 7x - 7 = 0 <=>
<=> x * (x + 1) - 7 * (x + 1) = 0 <=>
<=> (x + 1) * (x - 7) = 0 <=>
<=> x + 1 = 0 and x - 7 = 0 <=>
<=> x = -1 and x = 7, x ∈ { 6; +∞ } <=>
<=> x = 7
Good luck! :)
If you divide 5 by 6 to turn it into a decimal, you get 0.83. All you have to do is find a numerator that you can divide by 8 to get a decimal less than 0.83.
So, 1/8, 2/8, 3/8, 4/8, 5/8, or 6/8 would work.
Considering that the p-value associated for a r<em>ight-tailed test with z = 2.115</em> is of 0.0172, it is found that it is significant at the 5% level, but not at the 1% level.
<h3>When a measure is significant?</h3>
- If p-value > significance level, the measure is not significant.
- If p-value < significance level, the measure is significant.
Using a z-distribution calculator, it is found that the p-value associated for a r<em>ight-tailed test with z = 2.115</em> is of 0.0172, hence, this is significant at the 5% level, but not at the 1% level.
More can be learned about p-values at brainly.com/question/16313918
Please, Karla, explain what "C" and "n" represent. Are you talking about combinations (for example, n+2 objects taken n at a time? Or is C some kind of function? I don't quite see the relationship of this problem to 'data management.'
