Given the table showing the distance Randy drove on one day of her vacation as follows:
![\begin{tabular} {|c|c|c|c|c|c|} Time (h)&1&2&3&4&5\\[1ex] Distance (mi)&55&110&165&220&275 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7C%7D%0ATime%20%28h%29%261%262%263%264%265%5C%5C%5B1ex%5D%0ADistance%20%28mi%29%2655%26110%26165%26220%26275%0A%5Cend%7Btabular%7D)
The rate at which she travels is given by
If Randy has driven for one more hour at the same rate, the number of hours she must have droven is 6 hrs and the total distance is given by
distance = 55 x 6 = 330 miles.
Answer:
(13+83+85+87+90+91+93+97+98+99+100+100):12= 85,33
Answer:
Step-by-step explanation:
Just substitute in 7 for the n.
30-4(7)
30-28
2
The 7th term in the sequence is 2.
(please give brainliest?)
The answer is B. 8x+6. You multiple 3 and x+4 then combine like terms
Answer: The plane is 6,547 feet from the closer edge of the runway.
This is a trigonometry problem that will involve setting up 2 different triangles. One going to the closer edge of the runway and the other going to the far end of the runway.
The constant between the 2 triangles is the height of the plane. You can solve for this is both triangles and set them equal to each other.
Do that will give you the equation:
x/tan(70) = (x+3000)/tan(76) where x is the distance to the closer end of the runway.
If you solve that equation, you will get x = 6547 feet.
