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Dmitry_Shevchenko [17]
3 years ago
13

The function of c=p+. 07p relates the price of an item p with its cost c after 7% tax and s added. A student has $12.50. What is

the cost of the most expensive item he can buy?
Mathematics
1 answer:
disa [49]3 years ago
7 0

If you have $12.50 to spend, what is the cost of the most expensive item that you can buy? ... The sales tax in TN on nonfood items is 7%. .... Write a function rule to describe each situation The sellings price s(c) after a 45% markup ..... you want to write a program to calculate the total cost of an item that has sales tax added.


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A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
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Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

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Answer:

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