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slamgirl [31]
3 years ago
10

Which of the following uses the distributive property correctly?

Mathematics
1 answer:
Evgen [1.6K]3 years ago
5 0

Answer:

The last one is correct: 3(x+5)=3 . X + 3 . 5

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A halogen-lighting manufacturer packs 64 halogen lamps ink a cube-shaped container. the manufacturer has been asked by his distr
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64 divided by 8 = 8

So you're answer would be 8
4 0
3 years ago
If p is true and q is false, then p ∨ q is true. <br> A. True<br> B. False
Marysya12 [62]

Answer:

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Step-by-step explanation:

7 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
3 years ago
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