Answer:
it would be 30 because you can only use 1 side once so it would be 6 times 5
Answer:
D. about 8.5 mi
Step-by-step explanation:
To go from Aesha to Josh, you go 6 units right and 6 units up.
Each unit is a mile, so you go 6 miles right and 6 miles up.
Think of each 6 mile distance as a leg of a right triangle, and the direct distance from one place to the other as the hypotenuse of the right triangle. Use the Pythagorean theorem to find the length of the hypotenuse.
a^2 + b^2 = c^2
The 6-mile legs are a and b. c is the hypotenuse.
(6 mi)^2 + (6 mi)^2 = c^2
c^2 = 36 mi^2 + 36 mi^2
c^2 = 72 mi^2
c = sqrt(72) mi
c = sqrt(36 * 2) mi
c = 6sqrt(2) mi
c = 6(1.4142) mi
c = 8.5 mi
The answer is 18%
295-250. 45
————— = ———— = 18%
250. 250
The correct question is:
Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)
Answer:
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
Step-by-step explanation:
We need to verify that
x = c1e^(-t) + c2e^(3t)
is a solution to the differential equation
x''- 2x' - 3x = 0
We differentiate
x = c1e^(-t) + c2e^(3t)
twice in succession, and substitute the values of x, x', and x'' into the differential equation
x''- 2x' - 3x = 0
and see if it is satisfied.
Let us do that.
x = c1e^(-t) + c2e^(3t)
x' = -c1e^(-t) + 3c2e^(3t)
x'' = c1e^(-t) + 9c2e^(3t)
Now,
x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]
= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)
= 0
Therefore, the differential equation is satisfied, and hence, x is a solution.
