Let the train's original speed be s. Recall that distance = (speed)(time). Here, 150 miles = s(time), or 150 = s*t. If its original speed is increased by 5 mph, then the time required tomake the trip is 1 hour less than before;
distance = speed times time, so
150 mi = (s+5)(t-1). Let's eliminate t and solve for s:
Since 150 = s*t, t = 150/s. Subbing 150/s for t in the 2nd equation, we get:
150 s
150 mi = (s+5)(150/s - 1), or 150 = (s+5)(------- - ---- )
s s
Mult. both sides by s to elim. the fraction(s):
150s = (s+5)(150 - s)
Then 150s = 150s - s^2 + 750 - 5s, or
0 = -s^2 - 5s + 750
or 0 = s^2 + 5s - 750
Thus, 0 = (s-25)(s+30), and the roots are s=25 and-30. Only a positive original speed makes sense, so the answer is s = 25 mph.
Answer:
measure of what?
Step-by-step explanation:
Answer: +19
Step-by-step explanation: In this problem we are asked to evaluate 9 - (-12).
To subtract these integers, it's important to understand that when we have minus a negative, we can change it to plus a positive.
So 9 - (-12) can be changed to 9 + (+12) or 9 + 12 which gives us a sum of 21.
Therefore, 9 - (-12) = +21.
Notice that I put the '+' sign in front of the 21. Although this is optional, it's a good idea to do especially when adding and subtracting positives and negatives. All it shows is that our answer is positive.
Answer:
x = 16
Step-by-step explanation:
log₄ˣ = 2
x = 4²
x = 16
Step-by-step explanation:
Hope it will help you..