the correct question in the attached figure
(2^5)/8=2^2
we have
8----------- >2^3
(2^5)/8----------- > (2^5)/(2^3)=2^(5-3)=2^2
therefore
2^2=2^2-- ------> is ok
the answer is by simplifying 8 to 2^3 to make both powers base two and subtracting the exponents
Answer:
Step-by-step explanation:
<u>Complete the square</u>
<u />
You would subtract 15 on both sides and then you would have 2x=-14 then you divide both sides by 2 and x would equal -7
In this case, we cannot simply take the average speed by
adding the two speeds and divide by two.
What we have to do is to calculate the time required
going to school and the return trip home.
We know that to calculate time, we use the formula:
t = d / v
where,
d = distance = 4.8 km = 4800 m
v = velocity
Let us say that the variables related to the trip going
to school is associated with 1, and the return trip home is 2. So,
t1 = 4800 m / (22.6 m / s)
t1 = 212.39 s
t2 = 4800 / (16.8 m / s)
t2 = 285.71 s
total time, t = t1 + t2
t = 498.1 s
Therefore the total average velocity is:
= (4800 m + 4800 m) / 498.1 s
= 19.27 m / s = 19.3 m / s
Answer:
19.3 m/s
10, 0, -10
You just have to subtract by 10
