Volume of the sphere is 4/3*pi*r^3
here, r is 11 yd.
volume = 4/3*pi*11^3
= 4/3*pi*1331
= 5575.28 yd^3
= 5575.3 yd^3
Answer:
ITS 8 x 4 = 32 OK REEE
Step-by-step explanation:
Answer:
Using the formula:
where
A is the total amount
P is the principal
I is the Simple Interest
As per the statement:
Principal(P) = 5300 rupees.
rate of interest(r) = 6.5% = 0.065
Total amount(A) = 6678 rupees.
Then using above formula we have;
Subtract 5300 both sides we get;
or
We have to find the time period.
Using formula of Simple interest:
where r is the rate of interest (in decimal)
here, r = 6.5% = 0.065
Substitute the given values top find t:
⇒
Divide both sides by 344.5 we have;
Therefore, the time period t in years is 4 years
Answer:
y=5×+2
Step-by-step explanation:
5(1)+2=7
5(2)+2=12
5(3)+2=17
5(4)+2=22
Answer:
Tyler is correct. The temperature dropped at a rate of about 4° per hour between 4 and 6, while the temperature dropped at about 2.25° per hour between 6 and 10.
Edit: Explanation
The question is asking about which window of time had a <em>faster</em> decline in temperature, not a larger total change in temperature.
In a 2 hour timeframe, the temperature dropped 8°. (4-6 PM)
In a separate 4 hour timeframe, the temperature dropped 9°. (6-10 PM)
To find which window had a faster change in temp, I took the total temperature drop for each timeframe, then divided it by the number of hours each drop took.
8° / 2 = 4° per hour for 4-6 PM
9° / 4 = 2.25° per hour from 6-10 PM
Since the speed at which the temperature dropped per hour was greater from 4-6 PM than 6-10 PM, Tyler was correct.
