DE and EF are segments of the line DF, so they will add up to DF. This can be represented with this equation:
Combine like terms:
Add 15 to both sides:
Subtract 6x from both sides:
Divide both sides by 3 to get x by itself:
The value of x will be 8.
Plug this value of x into the expression for DF:
DF will equal 57.
The LCD of 1/2 and 5/6 is equal to the LCM of 2 and 6.
Multiples of 2 and 6
2: 2,4,6
6: 6,12,18
As you can see, the first number in both lines are 6.
Answer:
5.4 feet
Step-by-step explanation:
Let the reference line represents the sea level without any elevation.
At 1 P.M. the ocean surface was at an elevation of 2.2 feet along the line in the figure and the surf shop was 8.2 feet above of this elevated ocean's surface along the line in the figure.
Similarly, at 7 P.M. the ocean surface was at an elevation of -2.4 feet along the line in the figure and the deck in the restaurant was 8.2 feet above of this elevated ocean's surface along the line in the figure.
From the figure, the distance between the deck and the ocean's surface at 1 P.M.= The distance between the lines and .
=5.4 feet
Hence, the distance between the deck and the ocean's surface at 1 P.M. is 5.4 feet.
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)
720 because the 4 in 24 is lower than 5