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Kryger [21]
3 years ago
12

The expression 2.5a+2b gives the cost (in dollars) for a DVDs and b CDs at a moving sale. a. The cost for 1 DVD is $ and the cos

t for 1 CD is $ . b. The total cost for 4 DVDs and 4 CDs is $ . Question 2 c. Is $20 enough to buy 6 DVDs and 3 CDs?
Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer/Step-by-step Explanation:

Given:

a = no. of DVDs

b = no. of CDs

Total cost ($) for DVDs and CDs = 2.5a + 2b

a. The cost for 1 DVD = 2.5(a) = 2.5(1) = $2.5

The cost for 1 CD = 2(b) = 2(1) = $2

b. Total cost for 4 DVDs and 4 CDs:

Substitute a = 4, and b = 4 into the equation.

2.5(4) + 2(4)

10 + 8 = 18

Total cost for 4 DVDs and 4 CDs = $18

c. To find out if $20 would be enough to buy 6 DVDs and 3 CDs, substitute a = 6, and b = 3 into the equation. Solve for the total cost to see if it equals $20 or is less than $20. If it's greater than $20, then it won't be enough.

2.5(6) + 2(3)

15 + 6 = 21

6 DVDs and 3 CDs cost more than $20, therefore, $20 is not enough.

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2(6x4) + 2(6x2.4n) + 2(4x2.4n) what is the value of n
AleksandrR [38]

Hey there!

2(6 * 4) + 2(6 * 2.4n) + 2(4 * 2.4n) 

= 2(24) + 2(14.4n) + 2(9.6n)

= 48 + 28.8n + 19.2n

= 48 + 48n

= 48n + 48


Thus, your answer should be: 48n


Good luck on your assignment and enjoy your day!


~Amphitrite1040:)


6 0
2 years ago
If you’re using the inspection method, what constant term would you need in the numerator of this rational expression for (x+5)
Eva8 [605]
<span>If I am using the inspection method, the constant term that I would need in the numerator of this rational expression for (x+5) to be a common factor of (x^2+6x+14) / (x+5) is 11/(x+1)</span>
7 0
3 years ago
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I NEED THIS DONE ASAPPPP<br> expand 4(m+2)<br><br> 4(m+2) =
77julia77 [94]

4(m + 2) expanded is 4 x m and 4 x 2

simplified: 4m + 8

3 0
3 years ago
Solve 5c − c + 10 = 34. <br> −6 <br> 6 <br> 11 <br> −11
natima [27]

Subtract 10 and collect terms to get the variable term by itself on the left.

... 4c = 24

Divide by the coefficient of the variable, 4.

... c = 6

The appropriate choice is the second one:

... 6

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You don't actually have to sove the equation to find the right answer. You just need to see which answer works.

-6: 5(-6) -(-6) +10 = -14 ≠ 34

6: 5(6) -(6) +10 = 34 . . . . . . . this answer works (you can stop here)

11: 5(11) -(11) +10 = 54 ≠ 34

-11: 5(-11) -(-11) +10 = -34 ≠ 34

7 0
3 years ago
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
Thepotemich [5.8K]

Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

6 0
3 years ago
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