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seraphim [82]
2 years ago
7

Jarred sells DVDs. His inventory shows that he has a total of 3,500 DVDs. He has 2,342 more contemporary titles than classic tit

les. Let x represent the number of contemporary titles and y represent the number of classic titles. The system of equations models the given information for both types of DVDs.
x + y = 3,500

x – y = 2,342

Solve the system of equations. How many contemporary titles does Jarred have?
Mathematics
1 answer:
uysha [10]2 years ago
5 0

The number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

<h3>Simultaneous equation</h3>

Simultaneous equation is an equation which involves the solving for two unknown values at the same time.

  • number of contemporary titles = x
  • number of classic titles = y

x + y = 3,500

x – y = 2,342

Add both be equation

x + x = 3,500 + 2,342

2x = 5,842

x = 5,842 ÷ 2

x = 2,921

Substitute x = 2,921 into

x – y = 2,342

2,921 - y = 2, 342

-y = 2,342 - 2,921

-y = -579

y = 579

Therefore, the number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

Learn more about simultaneous equation:

brainly.com/question/17127685

#SPJ1

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A rectangle is four times as long as it is wide. If its length were diminished by 6 meters and its width were increased by 6 met
valkas [14]

Answer:

Length equals 16 and Width equals 4

Step-by-step explanation:

First let us create an equation. We can use L and W for length and width.

If the length is 4 times the width, then we end up with: L = 4W

It then says, " If its length were diminished by 6 meters and its width were increased by 6 meters, it would be a square."

Since a square has an equal length and width then we end up with:

L - 6 = W + 6

Knowing this we can just substitute the first equation into the second one leaving us with: 4W - 6 = W + 6

We then remove a W from both sides so that the right side is left with a 6, and add 6 to both sides to remove the -6 from the left one.

This leaves us with 3W = 12

W = 4, and if we put that into our first equation, L = 4W, then Length equals 16, and Width equals 4. We can check this by putting it into the 2nd equation. 16 - 6 = 4 + 6.

7 0
2 years ago
The area of the shaded region to the nearest tenth is __________.
goldfiish [28.3K]

Answer: 380 ft^2

Step-by-step explanation:

Length of sides of square=25

Diameter of semicircle=25

Radius=r=25/2=12.5

π=3.14

Area of square=length x length

Area of square=25 x 25=625

Area of semicircle=0.5 x π x r x r

Area of semicircle=0.5 x 3.14 x 12.5 x 12.5

Area of semicircle=245.3125

Area of shaded portion=area of square - area of semicircle

Area of shaded portion=625-245.3125

Area of shaded portion=379.6875

Area of shaded portion=380ft^2

approximately

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The vertices of quadrilateral PQRS are listed.
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A bag contains 10 marbles: 3 are green, 2 are red, and 5 are blue. Jose chooses a marble at random, and without putting it back,
Ludmilka [50]
We know that
g=3
r=2
b=5
total marbies=g+r+b------> 3+2+5----> 10

a) <span>probability that the first marble is red
P(red)=r/total marbies------------> 2/10-----> 1/5

b) </span><span>probability that the second marble is blue
in this case total marbles-------> 9
P(blue)=b/total marbles----------> 5/9

c) </span><span>the probability that the first marble is red and the second is blue
(1/5)*(5/9)=1/9

the answer is
1/9</span>
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4 years ago
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