<span>To estimate 586 - 321, you can round the numbers up or down to the nearest five.
586 can either become 570, rounding up, or 585, rounding down, which is closer. 321 becomes 320. It's not close enough to 325, nor to 315, for either of those to be good choices. 570 - 320 = 250 and 585 - 320 = 265. The actual solution to the original problem is 265, so, the second estimate, because they both go down by 1, is the closest.</span>
Answer:
(4x + 3) • (x + 2) • (3x - 2)
Thats finding the root polynomials hope it helps a bit
What is the question asking
Answer: 22.6875 miles
Step-by-step explanation:
Monday = 2 3/4 miles = 2.75 miles
Tuesday = 2.75 miles
Wednesday = 2.75 miles
Thursday = 2.75 miles
Friday = 2.75 miles
Saturday = 2.5 × 2.75 = 6.875 miles
Sunday = 0.75 × 2.75 = 2.0625 miles
Total = 2.75 miles + 2.75 miles + 2.75 miles + 2.75 miles + 2.75 miles + 6.875 miles + 2.0625 miles
Total = 22.6875 miles
Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be
4x + 2y = 20
There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.
When x=0,
4(0) + 2y = 20
y = 10
When x=1,
4(1) + 2y = 20
y = 8
When x=2,
4(2) + 2y = 20
y = 6
When x=3,
4(3) + 2y= 20
y = 4
When x = 4,
4(4) + 2y = 20
y = 2
When x = 5,
4(5) + 2y = 20
y = 0
When x = 6,
4(6) + 2y = 20
y = -2
A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:
Number of adult tickets Number of children tickets
0 10
1 8
2 6
3 4
4 2
5 0
