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3 years ago

12

Recreate the Unit circle below. Label in both degree and radians and identify the points (cosx, sinx). Are there any patterns th

at will help you remember all the points?

1 answer:

allsm [11]3 years ago

6 0

YES THEY ARE PATTERNS

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Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

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

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

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

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

4 years ago

Goran is riding his bicycle. He rides for 22.5 kilometers at a speed of 9 kilometers per hour. For how many hours does he ride?

astra-53 [7]

Answer: He rides for 2.5 hours or 2 and a half hours.

Step-by-step explanation:

He needs to complete a trajectory of 22.5 kilometers.

He rides 9 kilometers per hour.

Divide the total by the amount of kilometers that he rides.

$\frac{22.5}{9}=2.5h$

6 0

3 years ago

Order the numbers from least to greatest -3/4, 0.5, 2/3, -7/3, 1.2

BaLLatris [955]

-7/3, -3/4, 0.5, 2/3, 1.2

Brainliest please

5 0

2 years ago

A sequence is defined by the recursive formula f(n+1)=f(n)-2. If f(1)=18, what is f(5)?

N76 [4]

Answer:

$f(5)=10$

Step-by-step explanation:

A sequence is defined by the recursive formula $f(n+1)=f(n)-2$

If $f(1)=18,$ then

for $n=1: f(2)=f(1+1)=f(1)-2=18-2=16$

for $n=2: f(3)=f(2+1)=f(2)-2=16-2=14$

for $n=3: f(4)=f(3+1)=f(3)-2=14-2=12$

for $n=4: f(5)=f(4+1)=f(4)-2=12-2=10$

7 0

3 years ago

Read 2 more answers

What is the slope of a line that goes through the points (2, 5) and (8, 6) ?

jenyasd209 [6]

Answer:

m = 1 / 6

Step-by-step explanation:

Use this formula y2 - y1 / x2 -x1 to find the slope.

6 - 5 / 8 -2 = 1/6

8 0

3 years ago