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3 years ago

Is the relation shown in the arrow diagram a function? Explain. * (NEED HELP ASAP)

Mathematics

1 answer:

Paraphin [41]3 years ago

4 0

Answer:

It is a function.

Step-by-step explanation: Each input has only one output.

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The cost of carpeting a sqaure yard is $9.10. How much does it cost to carpet 12.5 square yards?

lesya692 [45]

Answer:

113.75$

Step-by-step explanation:

I. When a square yard = 9.10

so 1 * 9.1 = 9.1

if 12.5 square yard

then 12.5 * 9.1 = 113.75$

8 0

3 years ago

What is 1/3 of 240 dollars

faltersainse [42]

1/3 of 240 is 240/3 which is 80 dollars.

7 0

3 years ago

Read 2 more answers

I need help with problem

pychu [463]

Answer:

f(x)=4x(x-3)

Vertex=3/2, -9

Step-by-step explanation:

First, factorize it.

That is equal to 4x(x-3)

Therefore, x intercepts are 0 and 3, as at least one of them need to be 0.

To find the vertex, first find the x of the vertex, which you can do by finding the average of the x intercepts, so it will be 3/2

Then, find the y, by making x in the original equation equal 3/2. This will give us -9 as the y.

The vertex will by 3/2, -9

4 0

2 years ago

Miguel tells his teacher1/5 is the same as 20%. Which best justifies Miguel’s answer? a.5 goes into 100 twenty times, so 20% is

svetoff [14.1K]

Answer:

C. 5 goes into 100 twenty times, and 1 times 20 is 20

Step-by-step explanation:

Since, we know that when we multiply both numerator and denominator of a fraction by a same number then we obtain an equivalent fraction,

Here, the given fraction,

$\frac{1}{5}$

By the above statement,

$\frac{1}{5}=\frac{1\times 20}{5\times 20}=\frac{20}{100}$

Now,

$a\%=\frac{a}{100}$

$\implies \frac{20}{100}=20\%$

Hence,

$\frac{1}{5}=20\%$

Option C is correct.

3 0

3 years ago

Read 2 more answers

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