Option 3: a 90 degree rotation clockwise
You can tell that it is 90 degrees because the original started completely in quadrant 2 and the final image is completely in quadrant 1. If it was only rotated 45 degrees the final image would be part in quadrant 2 and part in quadrant 1. It was rotated clockwise because that is the way a clock goes.
Hope this helps! ;)
Answer:
Y= 2e^(5t)
Step-by-step explanation:
Taking Laplace of the given differential equation:
s^2+3s-10=0
s^2+5s-2s-10=0
s(s+5)-2(s+5) =0
(s-2) (s+5) =0
s=2, s=-5
Hence, the general solution will be:
Y=Ae^(-2t)+ Be^(5t)………………………………(D)
Put t = 0 in equation (D)
Y (0) =A+B
2 =A+B……………………………………… (i)
Now take derivative of (D) with respect to "t", we get:
Y=-2Ae^(-2t)+5Be^(5t) ....................... (E)
Put t = 0 in equation (E) we get:
Y’ (0) = -2A+5B
10 = -2A+5B ……………………………………(ii)
2(i) + (ii) =>
2A+2B=4 .....................(iii)
-2A+5B=10 .................(iv)
Solving (iii) and (iv)
7B=14
B=2
Now put B=2 in (i)
A=2-2
A=0
By putting the values of A and B in equation (D)
Y= 2e^(5t)
Answer:
B
Step-by-step explanation:
In a triangle, the sum of any two sides must be bigger than the third.
For the first one, 10+20=30 is not greater than 30, so this is not correct.
For B, 122+137 = 259 > 257, 257+137>122 , and 257 + 122 > 137. This works
For C, 8.6 + 2.7 = 11.3 < 12.2, so this does not work
For D, 1/6 + 1/5 = 5/(6*5) + 6/(6*5) = 5/30 + 6/30 = 11/30 < 1/2 = 15/30, so this does not work
