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3 years ago

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During a recent rainstorm, 888 centimeters of rain fell in Dubaku's hometown, and 2.362.362, point, 36 centimeters of rain fell

in Elliot's hometown. During the same storm, 15.1915.1915, point, 19 centimeters of snow fell in Alperen's hometown. How much more rain fell in Dubaku's town than in Elliot's town?

2 answers:

Elenna [48]3 years ago

7 0

Answer:

<em>5.64 centimeters more rain fell in Dubaku's town than in Elliot's town.</em>

Step-by-step explanation:

Amount of rainfall in Dubaku's hometown is 8 centimeters, amount of rainfall in Elliot's hometown is 2.36 centimeters and the amount of rainfall in Alperen's hometown is 15.19 centimeters.

So, <u>the difference of amount of rainfall for Dubaku's town and Elliot's town</u> will be: $(8- 2.36) centimeters = 5.64$ centimeters.

That means, 5.64 centimeters more rain fell in Dubaku's town than in Elliot's town.

guajiro [1.7K]3 years ago

3 0

Answer:

5.64

Step-by-step explanation:

i took the test on khan academy

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2 years ago

Which of the following functions are homomorphisms?

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Part A:

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

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

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

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 $f(x)=x$

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

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

Therefore, the function is a homomorphism.

Part C:

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

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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

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

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