All categories

3 years ago

What is the domain for the function represented by the graph?

Mathematics

2 answers:

EleoNora [17]3 years ago

6 0

Answer:

A. (-∞,+∞)

Step-by-step explanation:

Yuki888 [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis.

You might be interested in

Wht is the surface area of this design ?

soldi70 [24.7K]

2·(6·6+10·6+10·6)=312

4 0

3 years ago

Which of the following could be the system of nonlinear inequalities graphed below?

Ivan

Answer:

Option A.

Step-by-step explanation:

step 1

we know that

The equation of the solid line is

$y=5$

The solution is the shaded area above the solid line

The equation of the first inequality is

$y\geq 5$

step 2

The equation of the dashed line is

$y=x^{2} -5x+6$

The solution is the shaded area above the dashed line

The equation of the second inequality is

$y>x^{2} -5x+6$

therefore

The system of inequalities could be

$y\geq 5$

$y>x^{2} -5x+6$

4 0

3 years ago

Help me pls! PLS PLS PLS I NEED IT<br><br>20 points<br>options are x, x^9, x^4, and x^6

marishachu [46]

Answer:

x^4

Step-by-step explanation:

The expression can be simplified to (x^2)^2. Using an exponent rule, you multiply the two exponents together to get 4. The answer is x^4

5 0

3 years ago

Helena has taken out a $9,300 unsubsidized Stafford loan to pay for her college education. She plans to graduate in four years.

kherson [118]

The effective annual interest rate is:
i = (1 + 0.064/12)^12 - 1 = 0.066
In year 1: the interest is $613.80 (multiple $9300 by 0.066)
In year 2: the interest is $654.31 (add interest from year 1 to $9300 and multiply by 0.066)
In year 3: the interest is $656.98 (do the same as year 2)
In year 4: the interest is $657.16

The total interest is: $2582.25
The present worth of this amount is:
P = 2582.23 / (1 + 0.066)^4 = $1999.72

The answer is $1999.72.

8 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

4 years ago