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2 years ago

A line that is perpendicular to y=-4-10 passing through the point (8,11)

Mathematics

1 answer:

guapka [62]2 years ago

4 0

Answer:

Step-by-step explanation:

11=4(8)+b

11=32+b

-21=b

y=4x-21

i think

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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

3 years ago

Find the distance between y=4x+2,y=4x-3

KonstantinChe [14]

Pay attention in class and you would know

7 0

3 years ago

Given the numbers x = a and y = –b, which statement is true?

Minchanka [31]

choice D is correct, any number inside the absolute value sign is positive so choice A and C is eliminated, if there is a negative sign in front of the absolute value symbol, the number becomes negative which eliminates choice B. The only choice left is choice D

4 0

2 years ago

Read 2 more answers

Simplify this expression<br> (5x+2)^2

Galina-37 [17]

[ Answer ]

$\boxed{25x^2 \ + \ 20x\ + \ 4}$

[ Explanation ]

Simplify: $(5x \ + \ 2)^{2}$

------------------------------------------

Apply Perfect Square Formula ( $(a \ + \ b)^{2}$ = $a^{2}$ + 2ab + $b^{2}$ ) a = 5x, b = 2

$(5x)^{2}$ + 2 · 5x · 2 + $2^{2}$ : $25x^{2}$ + 20x + 4

Answer

$25x^{2}$ + 20x + 4

$\boxed{[ \ Eclipsed \ ]}$

5 0

2 years ago

What is X if X + 2/3 = 9/10?

laiz [17]

Answer:

$x=\frac{7}{30}$

Step-by-step explanation:

Given the following question:

$x+\frac{2}{3} =\frac{9}{10}$

In order to find the answer to this question we need to subtract 9/10 from 2/3. To do this we need to find the common denominator, and then subtracting the numerators.

$\frac{9}{10} -\frac{2}{3}$
$\frac{9}{10} \times3=\frac{27}{30}$
$\frac{2}{3} \times10=\frac{20}{30}$
$27-20=7=\frac{7}{30}$
$x=\frac{7}{30}$

Hope this helps.

4 0

2 years ago