Answer:
a = 2
Step-by-step explanation:
Let's find the equation of the line using the 2 points given.
Let's call -1 as x_1 and -5 as y_1
also, 4 as x_2 and 5 as y_2
The equation of line is :

<em>Let's plug the points and get the equation of the line:</em>

Now, to find a, we substitute a in x and 1 in y of the equation of the line we just got:

Answer:
a number 12
Step-by-step explanation:
duh bro
Part A:
Given

defined by


but

Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given

defined by

Note that in

, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular


and

Therefore, the function is a homomorphism.
Part C:
Given

, defined by


Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given

, defined by


but

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given

, defined by
![\left([x_{12}]\right)=[x_4]](https://tex.z-dn.net/?f=%5Cleft%28%5Bx_%7B12%7D%5D%5Cright%29%3D%5Bx_4%5D)
, where
![[u_n]](https://tex.z-dn.net/?f=%5Bu_n%5D)
denotes the lass of the integer

in

.
Then, for any
![[a_{12}],[b_{12}]\in Z_{12}](https://tex.z-dn.net/?f=%5Ba_%7B12%7D%5D%2C%5Bb_%7B12%7D%5D%5Cin%20Z_%7B12%7D)
, we have
![f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%2B%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Ba%2Bb%5D_%7B12%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%5Ba%2Bb%5D_4%3D%5Ba%5D_4%2B%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29%2Bf%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
and
![f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Bab%5D_%7B12%7D%5Cright%29%20%5C%5C%20%5C%5C%20%3D%5Bab%5D_4%3D%5Ba%5D_4%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29f%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
Therefore, the function is a homomorphism.
Answer:
1.49
Step-by-step explanation:
In order to find the slope of the tangent line to a given equation, and in a given point, we need to:
1. Find the first derivative of the given function.
2. Evaluate the first derivative function in the given point.
1. Let's find the first derivative of the given function:
The original function is 
But remeber that the derivative of
is 
so, 
2. Let's evaluate the first derivative function in the given point
The given point is (0.4,1.49) so:



Notice that the calculated slope of the tangent line is equal to the y-coordinate of the given point because f'(x)=f(x). In conclusion, the slope of the tangent line is equal to 1.49.
Answer:
that answer is 7 is bigger then 1.9
Step-by-step explanation:
so 7 is a whole and 1.9 is a decimal