Answer:
x = 2 ±i
Step-by-step explanation:
x^2 = 4x-5
Subtract 4x from each side
x^2 - 4x = 4x-5 -4x
x^2 - 4x= -5
Complete the square
Take the coefficient of the x term, divide by 2 and square it
-4/2 =2 2^2 =4
Add 4 to each side
x^2 -4x+4 = -5 +4
The left side is (x-coefficient of the x term/2)^2
(x-2)^2 = -1
Take the square root of each side
sqrt((x-2)^2) = sqrt(-1)
x-2 = ±i
Add 2 to each side
x-2+2 = 2 ±i
x = 2 ±i
The angular velocity is the linear speed divided by the radius.
ω = d/(t*r) = (14.13 ft)/((3 s)*(6 ft))
ω = 0.785 radians/second
There are 180/π degrees per radian, so the angular velocity can also be written as ...
ω = 45 degrees/second
1. Let a and b be coefficients such that
Combining the fractions on the right gives
so that
2. a. The given ODE is separable as
Using the result of part (1), integrating both sides gives
Given that y = 1 when x = 1, we find
so the particular solution to the ODE is
We can solve this explicitly for y :
![\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cln%5Cleft%7C%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%5Cright%7C)
![\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%7D)
2. b. When x = 9, we get
![y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B45%7D%7B21%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B15%7D7%7D%20%5Capprox%20%5Cboxed%7B1.29%7D)
Your going to want to first distribute the negative out to the second equation
65x+50y-16x+21y
49x+71y
Answer:
(You never showed the graphs we were suppose to choose from)
The graphs of all exponential functions have these characteristics. They all contain the point (0, 1), because a0 = 1. The x-axis is always an asymptote. They are decreasing if 0 < a < 1, and increasing if 1 < a.
