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3 years ago

Which ordered pairs are solutions to the equation?

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

6 0

To solve this you want to plug in your x's and y's to see if they match.

A. 3(11)-4=33-4=29 and 29 is not equal to 5 so A is not a solution

B.3(5)-4=15-4=11=11 3(3)-4=9-4=5 and 5 is not equal to 2 so B is not a solution

C. 3(2)-4= 6-4=2 2 is not equal to 3 so C is not a solution

D is the answer. 3(5)-4=11 and 3(2)-4= 2

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The stores markup on a wholesale item is 40%. The store is currently having a sale, and the item sales for 25% off the retail pr

Maurinko [17]

Answer:

75%

Step-by-step explanation:

7 0

3 years ago

There are eight employees on the Game Shop's sales teams. Last month, they sold a total of g games. One of the sales teams membe

liberstina [14]

Answer:

g/8 - 17

Step-by-step explanation:

Number of employees = 8

Total games sold = g

Average sales per employee = total games sold / number of employees = g/8

Chris sold 17 fewer games than the average sales per employee :

Chris sale = average sale per employee - 17

Chris sales = g/8 - 17

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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

3 years ago

Line m is drawn so that it is perpendicular to two distinct planes, Q and R. What must be true about planes Q and R?

UNO [17]

B. are parallel.

<span>Line m is drawn so that it is perpendicular to two distinct planes, Q and R. Therefore, planes Q and R are parallel.

Since it doesn't state that two planes intersect with each other, it is safe to assume that only line m is their connection. The planes are parallel with each other. </span>

7 0

3 years ago

Pls anyone help for a big change in my grade

gregori [183]

Answer:

x=3

Step-by-step explanation:

A vertical line will have the equation in terms of x. Any point you choose on line <em>l </em>will have an x-coordinate of 3.

6 0

2 years ago