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Alenkasestr [34]
3 years ago
5

What can you do when the unit prices for two comparable items are listed using different units of measure? a. The secondary unit

s don’t affect the cost. As long as both prices are in dollars or cents, you can compare the two. b. The unit prices are based on the cost of the whole package, so divide each base cost by each unit price to get comparable units. c. Use conversion rates to change one unit into the other unit, so that both prices are listed using like units. d. Unit prices are ratios, so you can add the two prices together and divide by the smaller unit price to get a usable number.
Mathematics
1 answer:
quester [9]3 years ago
6 0

Answer:

Answer is C on edge

Step-by-step explanation:

Trust me

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What is the value for x in the diagram? Rounded to the nearest whole<br> degree.
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Answer:

49

Step-by-step explanation:

cos(x)=31/47

x=cos^-1 (31/47)

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Write two different pairs of decimals whose sums are 8.69. one pair should involve regrouping.
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No regrouping: 4.34 + 4.35
Regrouping (carrying): 4.93 + 3.76
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in the balcony of a theatre there are 420 seats. the number of seats in each row is 14 more than the number of rows. find the nu
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420 divided by 14 = 30


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7 0
3 years ago
13+(w)/(7)=-18<br><br> step by step?
kotykmax [81]

w in (-oo:+oo)

w/7+13 = -18 // + 18

w/7+13+18 = 0

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3 0
3 years ago
A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha
Mariulka [41]

Answer:

V = 63π / 200  m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

                                 y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

                           S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx

- The derivative of the function f'(x) is as follows:

                            f'(x) = \frac{21-x}{\sqrt{42x-x^2} }

- The square of derivative of f(x) is:

                            f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }

- Now use the surface area formula:

                           S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

8 0
3 years ago
Read 2 more answers
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