Since the second line is parallel to the first, it will have the same slope. The y-intercept of the new line is 4 (because of the coordinates given) so the equation of Lynda's second cut is y = 1/4x + 4 which is choice C
Answer is <u>6|x+2|+2</u>
Please check the attached image of answer to clear the doubt.
<em>By </em><em>Benjemin</em><em> </em>☺️
Answer:
c=∞
Step-by-step explanation:
Well first step we can add 16+17 on both sides
33+c=33+c
-33 -33
c=c
The answer is infinity
Answer:
Between 15.95 ounces and 16.15 ounces.
Step-by-step explanation:
We have the following value m, being the mean, sd, being the standard deviation and n, the sample size:
m = 16.05
sd = 0.1005
n = 4
We apply the formula of this case, which would be:
m + - 2 * sd / (n ^ 1/2)
In this way we create a range, replacing we have:
16.05 + 2 * 0.1005 / (4 ^ 1/2) = 16.1505
16.05 - 2 * 0.1005 / (4 ^ 1/2) = 15.9495
Which means that 95% of all samples are between 15.95 ounces and 16.15 ounces.
Answer:
Step-by-step explanation:
(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.
(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.
(C) Example of a second order linear ODE:
M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)
The equation will be homogeneous if K(t)=0 and heterogeneous if 
Example of a second order nonlinear ODE:

(D) Example of a nonlinear fourth order ODE:
![K^4(x) - \beta f [x, k(x)] = 0](https://tex.z-dn.net/?f=K%5E4%28x%29%20-%20%5Cbeta%20f%20%5Bx%2C%20k%28x%29%5D%20%3D%200)