All categories

3 years ago

Article on sale original price percent discount

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

4 0

The 1/3% would equal .33% hopefully that helps at least a little

You might be interested in

The markup of an MP3 player is $207.20. The markup rate is 60%. What is the cost?

ch4aika [34]

If markup is 60% of the cost, you have

... markup = 0.60×cost

... markup/0.60 = cost

... $207.20/0.60 = cost = $345.33

_____

If markup is 60% of the selling price, the other 40% is the cost. So, the cost is 40/60 = 2/3 of the markup.

... cost = (2/3)×markup = (2/3)×$207.20

... cost = $138.13

_____

Markup can be specified in terms of either selling price or cost. Those are the usual reference values; there could be others. When only the percentage is given, you don't have enough information to work the problem. You need to know what it is a percentage of. (Example problems in your text may tell you the expected interpretation.)

5 0

3 years ago

If y=mx, and y is 12 when x is 4, then what is the constant of proportionality in this equation?A. 1/3B. 3C. 8D. 16

Anna007 [38]

$\begin{gathered} y=mx\text{ when x=4 and y=12} \\ 12=m(4) \\ 4m\text{ =12} \\ m=\frac{12}{4}=3 \end{gathered}$

The constant of proportionality,m, is 3 (option B)

5 0

1 year ago

At what point do the lines y = 6x – 18 and<br> y=8x – 32 intersect?

anzhelika [568]

Answer:

(7, 24)

Step-by-step explanation:

Solve the system:

6x - 18 = 8x - 32

2x = 14

x = 7

y = 6(7) - 18

y = 24

4 0

2 years ago

What is the closest whole number approximation of √50?

mr_godi [17]

Answer:

Step-by-step explanation:

√50 is close to √49

49 is a perfect square, 50 is close to 49.

√49 = 7

√50 ≈ 7.071068

6 0

2 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