Hello,
I am going to remember:
y'+3y=0==>y=C*e^(-3t)
y'=C'*e^(-3t)-3C*e^(-3t)
y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9
==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k
Answer:
The ant Fidencia ate more
Step-by-step explanation:
The question in English
The ant Fidencia found a leaf, the charge and took the anthill. She gave her friend 1/5 and she ate 1/3 of the leaf. Who ate more?
Her friend
Multiply (1/5) by (3/3)
(1/5)*(3/3)=3/15
Ant Fidencia
Multiply (1/3) by (5/5)
(1/3)*(5/5)=5/15
Compare
5/15 > 3/15 -----> two fractions with the same denominator the fraction with the highest numerator is greater
therefore
The ant Fidencia ate more
Answer:
106.1 ft/s
Step-by-step explanation:
You know the diagonal of a square is √2 times the length of one side, so the distance from 3rd to 1st is 90√2 feet ≈ 127.2792 ft.
The speed is the ratio of distance to time:
speed = distance/time = 127.2972 ft/(1.2 s) ≈ 106.1 ft/s.
_____
In case you have never figured or seen the computation of the diagonal of a square (the hypotenuse of an isosceles right triangle), consider the square with side lengths 1. The diagonal will cut the square into halves that are isosceles right triangles with leg lengths 1. Then the Pythagorean theorem can be used to find the diagonal length d:
d² = 1² + 1²
d² = 2
d = √2
Since this is the diagonal for a side length of 1, any other side length will serve as a scale factor for this value. A square with a side length of 90 ft will have a diagonal measuring 90√2 ft.
nmmn,.Answer:
mn,./.,m.mn,.mn,lm
Step-by-step explanation:
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