This is the answer: the x should be 52.5.
Answer: D. There are at least two oatmeal cookies next to each other.
Step-by-step explanation:
Alright so basically we can use process of elimination to determine what is true or false so
A. States the farthest to the left is chocolate but we can't prove that because all we know is the cookie farthest to the right is chocolate so this is<u> false</u>
B. Same reason as A we cannot prove what cookie is farthest to the left because we are not given a pattern so <u>B is false</u>
C. Since there are definitely twice as many girls as boys in the class and that also means there are twice as many oatmeal cookies then we cannot prove that 2 chocolate cookies have to be next to each other<u> so also false</u>
<u></u>
D. This has to be true because if there are twice as many girls as boys and more oatmeal than chocolate then is whatever cookie line combinataion we will have at least 2 oatmeal cookies next to each other So this is true
E. This is most definitely not true because the question tells us that twice as many girls gave him oatmeal so if anything there are more oatmeal cookies
If the fruit salad must contain watermelon, he can make the fruit salad in 35 ways.
Let's start with day 1:
We know that Joe practiced from 6:30 to 7:05
An hour is 60 minutes
Joe practiced for the 30 minutes left in that hour and 5 minutes into the next
30+5=35
So, he was playing for 35 minutes on the first day
Now, we can do the same thing for the second day:
He started at 3:55 and ended at 4:15
An hour is 60 minutes
<span>Joe practiced for the 5 minutes left in the hour and the first 15 minutes of the next hour
</span>
5+15=20
Now we want to add the number of minutes he practiced for both days
35+20=55 minutes
Answer:
x≤-3 2 ≤x<7
Step-by-step explanation:
(x^2+x-6) /(x-7) ≤0
Factor the numerator
(x+3)(x-2) / (x-7)≤0
Critical points are
X =-3 X =2 X=7
We need to check the values when
x≤3 -3 ≤ x≤2 2 ≤x<7 x>7
(x cannot be 7 because then the denominator is zero)
and see when it is less than 0
Substitute a value in and see if factor is positive or negative
x≤-3
(negative)(negative) / (negative) = negative
-3 ≤x≤2
(positive)(negative) / (negative) = positive this range does not work
2 ≤x<7
(positive)(positive) / (negative) = negative
x>7
(positive)(positive) / (positive)