!!!!!!!50 points!!!!!!!! QUESTION BELLOW HELP ASAP!!!!!!!!!
Answers
Answer 1
Answer:
hmm, interesting
one way would be to multply it out or set it equal to 11x where x is a whole number (if x is not a whole number, then it is not divisible)
11x=7^6+7^5-7^4
undistribute 7^4
11x=(7^4)(7^2+7^1-1)
11x=(7^4)(49+7-1)
11x=(7^4)(55)
56=5*11
11x=(7^4)(5)(11)
divide by 11
x=5(7^4)
aka, find if 11 is a factor of that number
x=5(7^4)
one way would be to multply it out or set it equal to 11x where x is a whole number (if x is not a whole number, then it is not divisible)
11x=7^6+7^5-7^4
undistribute 7^4
11x=(7^4)(7^2+7^1-1)
11x=(7^4)(49+7-1)
11x=(7^4)(55)
56=5*11
11x=(7^4)(5)(11)
divide by 11
x=5(7^4)
aka, find if 11 is a factor of that number
x=5(7^4)
Answer 2
Answer:
Well for this question you can factor like this:
7^3(7^3 + 7^2 -7) = 7^3(385) = 7^3(35)(11) = 7^3(7)(5)(11) = 7^4(5)(11) :)))
I hope this is helpful
have a nice day
7^3(7^3 + 7^2 -7) = 7^3(385) = 7^3(35)(11) = 7^3(7)(5)(11) = 7^4(5)(11) :)))
I hope this is helpful
have a nice day
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Answers
Wouldn't it be 1? Because any number to the zero post power is 1?
Answers
the cattle train
speed - 36.5 km/h
time - x h
the passenger train
speed - 29.2 km/h
time - (2.4+x) h
The cattle train caught up to the passenger train so the distance they travelled is the same.
Distance is speed times time.
%20%5C%5C%2036.5x%3D70.08%2B29.2x%20%5C%5C%2036.5x-29.2x%3D70.08%20%5C%5C%207.3x%3D70.08%20%5C%5C%20x%3D%5Cfrac%7B70.08%7D%7B7.3%7D%20%5C%5C%20x%3D9.6%20%5C%5C%20%5C%5C%202.4%2Bx%3D2.4%2B9.6%3D12)
The passenger train travelled for 12 hours before the cattle train caught up.
speed - 36.5 km/h
time - x h
the passenger train
speed - 29.2 km/h
time - (2.4+x) h
The cattle train caught up to the passenger train so the distance they travelled is the same.
Distance is speed times time.
The passenger train travelled for 12 hours before the cattle train caught up.
Used brute force to figure this one out and got lucky at t=12.
What you need to know:
t = time travelled (in hours)
x = distance passenger train has travelled
y = distance cattle train has travelled
d = distance between both trains in km
------------------------------------
RESULTS AND TIME SNAPSHOTS:
When:
t=0, x=0, y=0, d=0
t=2.4, x=70.08, y=0, d=70.08
t=4.8, x=140.16, y=87.6, d=52.56
t=7.2, x=210.24, y=175.2, d=35.04
t=9.6, x=280.32, y=262.8, d=17.52
t=12, x=350.4, y=350.4, d=0 <-------
So your answer is 12 hours.
What you need to know:
t = time travelled (in hours)
x = distance passenger train has travelled
y = distance cattle train has travelled
d = distance between both trains in km
------------------------------------
RESULTS AND TIME SNAPSHOTS:
When:
t=0, x=0, y=0, d=0
t=2.4, x=70.08, y=0, d=70.08
t=4.8, x=140.16, y=87.6, d=52.56
t=7.2, x=210.24, y=175.2, d=35.04
t=9.6, x=280.32, y=262.8, d=17.52
t=12, x=350.4, y=350.4, d=0 <-------
So your answer is 12 hours.
Answers
Answers
No, because some angles are obtuse or acute....
Some are not because some are 180 degrees
Answers
6^2*4=36*4=144
Hope it was helpful
Hope it was helpful
6^2 is equivalent to 6*6. The problem written out would be:
6*6*4
36*4
144
6*6*4
36*4
144
Answers
6 1/2 Is your answer