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1 year ago

NEED HELP 10 points

Mathematics

2 answers:

meriva1 year ago

7 0

You said 10 points ???2

Marysya12 [62]1 year ago

6 0

Answer:

Original: 2 x 2 + 3^2 - 4 + 3 * 5

Answer: (2 * 2 + 3)^2 - [(4 + 3) * 5]

Step-by-step explanation:

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Choose the problem that has the same answer as 6 × 2 × 3.

11Alexandr11 [23.1K]

Answer:

6x3x2

Step-by-step explanation:

8 0

3 years ago

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

Write the equation log216 36=36 in logarithmic form

Zielflug [23.3K]

This is impossible and it is not even an equation since $\log_{216} 36 = 2/3$

3 0

3 years ago

Write the equation of the line that passes throught the point (-4,-9) and has a slope of 1/2

Ray Of Light [21]

Answer: y = 1/2x - 7

Data

Point = (-4, -9)

slope = m = 1/2

Process

To solve this problem, use the formula slope- y-coordinate, substitute the values and simplify.

Equation

y - y1 = m (x - x1)

x1 = -4

y1 = -9

Substitution

y + 9 = 1/2 (x + 4)

Simplification

y + 9 = 1/2x + 4/2

y = 1/2x + 2 - 9

Answer

y = 1/2x - 7

8 0

3 years ago

What is the equation of this line? y=1/2x-3 y=-1/2x-3 y=-2x-3 y=2x-3

quester [9]

Answer:

y=1/2x-3

Step-by-step explanation:

y=mx+b where m equals the slope and b equals the y-intercept. The slope is 1/2 and the y-intercept is -3. Hope this helps!

6 0

3 years ago

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