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zhuklara [117]
4 years ago
6

A jet flies over the Air Force Base at 10:20 a.m. At 10:32 a.m., the plane passes over the Navy Base Station, 120 miles away. Ho

w fast is the aircraft traveling?
Mathematics
1 answer:
kati45 [8]4 years ago
3 0
So first calculate the time between the plane flied over the Air Force Bases (AFB) and Navy Base Station (NBS):
(time passed Air Force Bases) - (time passed Air Force Bases) = 10:32 - 10:20 = 12 mins
Since we know the distance that plane travel from AFB to NBS is 120 miles and the plane traveled that distance within 12 minutes.
We have
120 miles / 12 minutes = 10 miles / minute
Calculate it into miles per hour (60 minutes):
10 miles/minute x 60 minutes = 600 miles per hour.
ANSWER : 600 mph
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3 years ago
There are 10 boys and 12 girls in the tennis club. The coach wants to select two players to practice first. Which statements are
Sidana [21]
Since there are 22 participants all in all. The possible combinations of the two picked for practice first is, 
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3 years ago
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The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=
Leno4ka [110]
1 step: S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}, S_{3}=T_{1}+T_{2}+T_{3}, then
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2 step: T_{n}=T_{1}*q^{n-1}, then 
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T_{5}=T_{1}*q^{4}
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and \left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right. will have form \left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right..

3 step: Solve this system  \left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right. and dividing first equation on second we obtain \frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}. So, \frac{(q-1)(q+1)}{q+1} = \frac{1}{2} and q-1= \frac{1}{2}, q= \frac{3}{2} - the common ratio.

4 step: Insert q= \frac{3}{2}into equation T_{1}*q^{3}*(q+1)=5 and obtain T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5, from where T_{1}= \frac{16}{27}.




5 0
3 years ago
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