To solve this you want to plug in your x's and y's to see if they match.
A. 3(11)-4=33-4=29 and 29 is not equal to 5 so A is not a solution
B.3(5)-4=15-4=11=11 3(3)-4=9-4=5 and 5 is not equal to 2 so B is not a solution
C. 3(2)-4= 6-4=2 2 is not equal to 3 so C is not a solution
D is the answer. 3(5)-4=11 and 3(2)-4= 2
Answer:
The answer is 21 minutes
Step-by-step explanation:
We use the equation Xf = Xo + vt
1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])
2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]
3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]
- Child 1 final position =>
- Child 2 final position =>
4) Sum the equations and equate to 3
5) Substitute the values we already know
6) in 15 + 6 minutes they will be 3miles apart
7) In 21 minutes they will still be able to communicate with one another.
2√72 / √8 ± √2
If the denominator is √8 + √2,
2√72 / √8 + √2
= 2√(2*2*2*3*3) / √(2*2*2) + √2
= 2*2*3√2 / 2√2 + √2
= 12√2 / 3√2
= 4
F(2)=2(3*2)+1
f(2)=2(6)+1
=13
answer is 13