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2 years ago

Find the inverse: f(x)= x+2/3

Mathematics

1 answer:

Kobotan [32]2 years ago

4 0

I’m not sure if this is right but f(x) = 1/x + 3/2 ?

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Helppp!!!!! 2 questions answer both

yKpoI14uk [10]

1. The answer would be about 20.72, or 20.7142 if you don’t want the rounded version

2. The answer is -12.8

3 0

2 years ago

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Solve the following: - 3 raised to 1 by 5 the whole raised to 4 (3^1/5)^4

Diano4ka-milaya [45]

Answer:

8.30256

Step-by-step explanation:

Step 1: Write out expression

$((-3)^{\frac{1}{5} })^{4(3^{\frac{1}{5} })^4$

Step 2: Use BPEMDAS to evaluate

$(-1.24573)^{4(3^{\frac{1}{5} })^4$

$(-1.24573)^{4(1.24573)^4$

$(-1.24573)^{4(2.40822)$

$(-1.24573)^{9.6329}$

= 8.30256

And we have our answer!

8 0

2 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

2 years ago

If y varies inversely as x, and y=6 when x=3. Then find the value of y when x=32

s2008m [1.1K]

Answer: y = 18/32

Step-by-step explanation: Okay the equation is xy = k

You plug in x and y.

(3)(6) = k

18 = k

To get the value of y when x = 32 you plug in 32.

Use this equation now y = k/x

Now it is y = 18/32

And that is your answer!

7 0

2 years ago

There are 16 ounces in one pound.

Kobotan [32]

Answer:

Step-by-step explanation:

You just multiply 16 by 4.

7 0

2 years ago

Read 2 more answers