Answer:
Step-by-step explanation:
y = -(x - 6)² + 2
At x-intercept y = 0
0 = -(x - 6)² + 2
(x - 6)² = 2
Take square root both sides
x - 6 = ± √2
x = 6 ± √2
4y - 3 = y + 6
Add 3 to both sides
4y = y + 9
Subtract y from both sides
3y = 9
Divide both sides by 3
y= 3
Hello there!
First, let's establish what the greatest common factor means.
The greatest common factor is the highest factor in value that's in comparison to all numbers.
For example;
2, 4, and 8. Find the GCF.
2 = 1 x 2
4 = 1 x 4, 2 x 2
8 = 1 x 8, 2 x 4
These numbers all have 2 as the highest number in common, so 2 is the GCF in this example.
Now, let's try with our current numbers.
30 = 1 x 30, 3 x 10, 5 x 6
48 = 1 x 48, 2 x 24, 3 x 16, 4 x 12, 6 x 8
60 = 1 x 60, 2 x 30, 3 x 20, 4 x 15, 5 x 12, 6 x 10
All of our numbers have these common factors:
1, 2, 3, 5, 6.
The greatest number is 6, so our GCF is 6.
I hope this helps!
Answer:
Yes, because the bike order meets the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100
Step-by-step explanation:
4c + 6a ≤ 120:
You can set up the inequality 4c + 6a ≤ 120, because it takes 4 hours to build per child bike (c) and 6 hours to build an adult bike (a), all together this time cannot surpass 120 hours. That is why you use the 'less than or equal to' sign.
4c + 4a ≤ 100:
You can then set up the inequality 4c + 4a ≤ 100, because it takes 4 hours to test a child bike (c) and 4 hours to test an adult bike (a). Since 100 hours is the max amount of time they can use to test out bikes, you will use the 'less than or equal to' sign.
4(5) + 6(15) = 20 + 90 = 110. 110 is less than 120
4(5) + 4(15) = 20 + 60 = 80. 80 is less than 100
Yes, because the bike order meets the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100
Answer:
B. C. D are true, A. is false
Step-by-step explanation:
<em>The question is attached</em>
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<u>Answer options:</u>
A. Players at school 1 typically spent more time in the weight room than players at school 2
- False, as per plot boxes up to 75% of players from school 1 spend up to 10 hours but up to 75% of players from school 2 spend up to 12 hours in the weight room
B. The middle half of the data for school 1 has more variability than the middle half of the data for school 2
- True. 4 hours variation for school 1 vs 3 hours variation for school 2
C. The median hours spent in the weight room for school 1 is less than the median for school 2 and the interquartile ranges for both schools are equal
- True as it is 8 hours for school 1 and 9 hours for school 2
D. The total number of hours spent in the weight room for players at school 2 is greater than the total number of hours for players at school 1
- True. Taking into account a sum of 5 numbers of the plot box. For School 1 it is: (3+4+8+10+16)= 41 and for school 2 it is: (3+6+9+12+14) = 44. Totals will have similar ratio.