Answer:
8x
Step-by-step explanation:
Because you are adding the (x+3) there you can immediately you the association property to simplify
(2x + 5x + x) + (3 - 3)
8x + 0 =
8x
Answer:
R(x) =300·x - 2·x²
C(x) = £5000 + £40 × x
The break even points are 23.47 and 106.53 or 23 and 107 bikes
Step-by-step explanation:
Given that the price function P(x) = 300 -2·x
Cost per bike = £40
The revenue function R(x) is given by bike price × total number of bikes manufactured and sold
∴ R(x) = P(x)×x = (300 - 2·x)×x = 300·x - 2·x²
The company's cost function, C(x) is Fixed cost + cost to produce each bike × total number of bikes produced
∴ C(x) = £5000 + £40 × x
The break even point is given by the relation;
Total revenue - total cost = 0
That is, break even point is R(x) - C(x) = 0
300·x - 2·x² - (5000 + 40·x) = 0
-2·x²+260·x-5000 = 0 or 2·x²- 260·x + 5000 = 0
Factorizing, we have;
(x - (65 -5√69))(x - (65 +5√69))
Solving gives x = 23.47 or 106.53
Therefore, the break even points are 23.47 and 106.53.
That is the company is profitable when they produce less than 23 bikes or more than 107 bikes.
Answer:
picture is too small .i am not able to see
You can pick a symbol, maybe m or z to be the symbol.
Answer:
The radius of a sphere is 2 millimeters
The surface area of a sphere is 50.24 square millimeters.
The circumference of the great circle of a sphere is 12.56 millimeters.
Step-by-step explanation:
<u><em>Verify each statement</em></u>
case A) The radius of a sphere is 8 millimeters
The statement is false
we know that
The volume of the sphere is equal to
![V=\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
we have
![V=100.48/3\ mm^{3}](https://tex.z-dn.net/?f=V%3D100.48%2F3%5C%20mm%5E%7B3%7D)
![\pi =3.14](https://tex.z-dn.net/?f=%5Cpi%20%3D3.14)
substitute and solve for r
![100.48/3=\frac{4}{3}(3.14)r^{3}](https://tex.z-dn.net/?f=100.48%2F3%3D%5Cfrac%7B4%7D%7B3%7D%283.14%29r%5E%7B3%7D)
![r^{3}=(100.48)/(4*3.14)](https://tex.z-dn.net/?f=r%5E%7B3%7D%3D%28100.48%29%2F%284%2A3.14%29)
case B) The radius of a sphere is 2 millimeters
The statement is True
(see the case A)
case C) The circumference of the great circle of a sphere is 9.42 square millimeters
The statement is false
The units of the circumference is millimeters not square millimeters
The circumference is equal to
![C=2\pi r](https://tex.z-dn.net/?f=C%3D2%5Cpi%20r)
we have
![\pi =3.14](https://tex.z-dn.net/?f=%5Cpi%20%3D3.14)
substitute
![C=2(3.14)(2)](https://tex.z-dn.net/?f=C%3D2%283.14%29%282%29)
![C=12.56\ mm](https://tex.z-dn.net/?f=C%3D12.56%5C%20mm)
case D) The surface area of a sphere is 50.24 square millimeters.
The statement is True
Because
The surface area of the sphere is equal to
![SA=4\pi r^{2}](https://tex.z-dn.net/?f=SA%3D4%5Cpi%20r%5E%7B2%7D)
we have
![\pi =3.14](https://tex.z-dn.net/?f=%5Cpi%20%3D3.14)
substitute
![SA=4(3.14)(2)^{2}](https://tex.z-dn.net/?f=SA%3D4%283.14%29%282%29%5E%7B2%7D)
![SA=50.24\ mm^{2}](https://tex.z-dn.net/?f=SA%3D50.24%5C%20mm%5E%7B2%7D)
case E) The circumference of the great circle of a sphere is 12.56 millimeters.
The statement is true
see the case C
case F) The surface area of a sphere is 25.12 square millimeters
The statement is false
because the surface area of the sphere is ![SA=50.24\ mm^{2}](https://tex.z-dn.net/?f=SA%3D50.24%5C%20mm%5E%7B2%7D)
see the case D