A function assigns the values. The cost of 6 CDs will be $32.
<h3>What is a Function?</h3>
A function assigns the value of each element of one set to the other specific element of another set.
Given that the Recycled CDs. Incorporated, offers a choice of 5 used CDs for $27, with each additional CD costing $5. Therefore, a cost function for purchasing 5 or more CDs, where x represents the number of CDs over 5 can be written as,
C(x) = $27 + $5(x)
Now, if 6 CDs are purchased then the number of CDs that are more than 5 is 1. therefore, the cost of 6 CDs will be,
C(1) = $27 + $5(1)
= $27 + $5
= $32
Hence, the cost of 6 CDs will be $32.
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*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*
(21)
Area of a Regular Hexagon:
square units
(22)
Similar to (21)
Area =
square units
(23)
For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:
![altitude=\frac{\sqrt{3}}{2}*side](https://tex.z-dn.net/?f=altitude%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2Aside)
![side = \frac{36}{\sqrt{3}}](https://tex.z-dn.net/?f=side%20%3D%20%5Cfrac%7B36%7D%7B%5Csqrt%7B3%7D%7D)
Hence, area of the hexagon will be:
square units
(24)
Given is the inradius of an equilateral triangle.
![Inradius = \frac{\sqrt{3}}{6}*side](https://tex.z-dn.net/?f=Inradius%20%3D%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B6%7D%2Aside)
Substituting the value of inradius and calculating the length of the side of the equilateral triangle:
Side = 16 units
Area of equilateral triangle =
square units
Answer:
A group of ten people would be willing to spend 135$
tell me if im wrong plz
Step-by-step explanation:
Answer:
6541.7 mm³ (nearest tenth)
Step-by-step explanation:
![\boxed{volume \: of \: cone = \frac{1}{3}\pi {r}^{2}h }](https://tex.z-dn.net/?f=%5Cboxed%7Bvolume%20%5C%3A%20of%20%5C%3A%20cone%20%3D%20%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%7Br%7D%5E%7B2%7Dh%20%20%7D)
Given that the height is 40mm, h= 40mm.
Diameter= 2(radius)
Radius of cone
= 25 ÷2
= 12.5mm
Volume of cone
![= \frac{1}{3} (3.14)(12.5)^{2} (40)](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B1%7D%7B3%7D%20%283.14%29%2812.5%29%5E%7B2%7D%20%2840%29)
= 6541.7 mm³ (nearest tenth)