Let's look at each one of the intervals.
Interval A
Interval A looks like she's going at a pretty steady pace, so Interval A is not the correct answer.
Interval B
Interval B looks like she stays at 2.5 for an hour. That looks like a traffic jam to me!
The correct answer is Interval B.
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Answer: He actually rode 2 miles per hour on his trip
Step-by-step explanation: Maybe unconventional, but express the time it took, then figure the speed.
Time = distance /speed t will represent time, s is the speed: t = 30/s Use the rime it would have taken at the higher speed to create an equation:
t-12 = 30/s+8 replace the y with the 30/s
30/s -12 = 30/s+8
(s)(30/s -12 ) = (s)(30/s+8 ) Cross multiply to cancel denominators
(s-8)(30 -12s) = (s-8)(30s/s+8 ) ==> 30s +240 -12s² -96s =30s Simplify:
(-1)(-12s² -96s +240 ) =0 ==> 12s² +96s -240 divide all by 12
s² + 8s -20 = 0 Factor and solve for s
(s +10)(s -2) =0 s-2=0 S= 2
Proof:
30/2 = 15 hours for original trip at 2mph,
increase speed by 8mph 2 + 8 = 10mph
30 miles at 10mph takes 3 hours; that is 12 hours less than his actual trip.
(Brainilest, please :-)
Answer:
She answered 36 questions correctly out of the 45 questions.
A = l x w
so you know that the length is 3m longer than the width, so you could use a formula to represent that
w = l + 3
you then substitute the second equation into the first to solve for l
70 = l x (l +3)
70 = l^2 + 3l
you could then rearrange the formula and solve for l using the quadratic formula
0 = l^2 + 3l - 70
l = -3 +- (square root (3)^2 - 4(1)(70)) / 2(1)
l = -3 +- (square root 9 + 280) / 2
l = -3 +- (square root 289) / 2
l = -3 +- 17 / 2
then you solve for the two seperate roots
l = -3 + 17 /2
l = 14 / 2
l = 7
or
l = -3 - 17 / 2
l = -20 / 2
l = -10
since a length cannot be negative, this root is not viable. therefore l = 7
to solve for w you would use
w = l + 3
w = 7 + 3
w = 10
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