The sign shows distances from a rest stop to the exits for different towns along a straight section of highway. The state depart
ment of transportation is planning to build a new exit to Freestone at the midpoint of the exits for Roseville and Edgewood. When the new exit is built, what will be the distance from the exit for Midtown to the exit for Freestone?
Midtown 17 mi.
Roseville 33 mi.
Edgewood 55 mi.
vertze vertze
When the new exit is built the distance from the exit for Midtown to the exit for Freestone will be miles.
Midtown 17 mi.
Roseville 33 mi.
Edgewood 55 mi.
vertze vertze
When the new exit is built the distance from the exit for Midtown to the exit for Freestone will be miles.
1 answer:
Refer to the diagram shown below.
The exit for Freestone is built midway between Roseville and Edgewood,
therefore the distance from O to the new exit is
(1/2)*(33+55) = 44 mi.
Let x = distance from Midtown to the new exit.
Because the distance from O to the new exit is equal to (x + 17), therefore
x + 17 = 44
x = 44 - 17 = 27 mi.
Answer:
When the new exit is built, the distance from the exit for Midtown to the exit for Freestone will be 27 miles.
The exit for Freestone is built midway between Roseville and Edgewood,
therefore the distance from O to the new exit is
(1/2)*(33+55) = 44 mi.
Let x = distance from Midtown to the new exit.
Because the distance from O to the new exit is equal to (x + 17), therefore
x + 17 = 44
x = 44 - 17 = 27 mi.
Answer:
When the new exit is built, the distance from the exit for Midtown to the exit for Freestone will be 27 miles.
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Answer:
There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.
The needed function is
Step-by-step explanation:
Let us assume the piece of paper James decides to fold is a SQUARE.
Now, let us assume:
n : the number of times the paper is folded.
y : The number of parts obtained after folds.
Now, if the paper if folded ONCE ⇒ n = 1
Also, when the pap er is folded once, the parts obtained are TWO equal parts.
⇒ for n = 1 , y = 2 ..... (1)
Similarly, if the paper if folded TWICE ⇒ n = 2
Also, when the paper is folded twice, the parts obtained are FOUR equal parts.
⇒ for n = 2 , y = 4 ..... (2)
⇒
Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7
So,
⇒ There are total 128 equal parts.
Lastly, if the paper is folded ELEVEN times ⇒ n = 11
So,
⇒ There are total 2048 equal parts.
Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.
And the needed function is