Answer:
We have the problem:
"Clare is in charge of getting snacks for a road trip with her friends and her dog. She has
$35 to go to the store to get some supplies. The snacks for herself and her friends cost
$3.25 each, and her dog's snacks costs $9 each."
In this situation we have two variables:
X = number of snacks for herself and her friends that she buys. Each one of these costs $3.50
Y = number of snacks for her dog that she buys. Each one of these costs $9.
The total cost, in this case, can be written as:
X*$3.50 + Y*$9
And we know that she has $35 to spend, so she can spend $35 or less in the store, then we have the inequality:
X*$3.50 + Y*$9 ≤ $35
Where we defined all the quantities in the inequality.
Hello!
Given the two points,
and
, and to find the distance between these two points is found by using the formula:

is assigned to one the points, in this case, is (4, 1).
is assigned to other point, which is (9, 1).
Then, plug in these values into the formula and solve.




Therefore, the distance between the two points is 5.


solve for "l" to find its length
Answer:
the 20 +2
Step-by-step explanation:
D) because the final answer should be 8/3. See the attachment.