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myrzilka [38]
3 years ago
15

Carl is boarding a plane. He has 2 checked bags of equal weight and a backpack that weighs 4 kg. The total weight of Carl's bagg

age is 35 kg.
Write an equation to determine the weight, w, of each of Carl's checked bags.
Mathematics
2 answers:
aniked [119]3 years ago
5 0
<span>Carl has 3 bags in total. One backpack weighs 4 kg and the rest two checking bags have the equal weight. The total weight of 3 bags is given to be 35 kg.

Let the weight of each checking bag is w kg. So we can write:

2 x (Weight of a checking bag) + Weight of Backpack = 35

Using the values, we get:

2w+ 4 = 35

Using this equation we can find the weight of each checking bag, as shown below.

2w = 31

w = 31/2

w = 15.5

Thus, the weight of each checking bag is 15.5 kg
</span>
Len [333]3 years ago
4 0

Answer: Weight of each of Carl's checked bags is 15.5 kg.

Step-by-step explanation:

Since we have given that

Number of checked bags of equal weight = 2

Weight of a backpack = 4 kg

Total weight of Carl's baggage = 35 kg

Let the weight of checked bags be 'w'.

According to question,

2w+4=35\\\\2w=35-4\\\\2w=31\\\\w=\frac{31}{2}\\\\w=15.5\ kg

Hence, Weight of each of Carl's checked bags is 15.5 kg.

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Which of the following is a perfect square trinomial?
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A.

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Take a look at what happens when squaring either of these...

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Notice a couple of patterns.

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2.  The coefficient of the middle term is either +2ab \text{ or } -2ab .  So what are <em>a</em>  and <em>b</em>?  <em>a</em> is the square root of the x^2 term and <em>b</em> is the square root of the y^2 term.

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Adam is building a computer desk with a separate compartment for the computer. The compartment for the computer is a rectangular
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Answer:

The width of the computer compartment is 9 inches

The length of the computer compartment is 33 inches

The height of the computer compartment is 27 inches

Step-by-step explanation:

The given data on the rectangular prism compartment are;

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The length of the compartment, l = 24 inches + The width, w

The height of the compartment, h = 18 inches + The width, w

Therefore, we have;

l = 24 + w

h = 18 + w

V = l × h × w

∴ V = (24 + w) × (18 + w) × w = w³ + 42·w² + 432·w = 8019

w³ + 42·w² + 432·w - 8019 = 0

By graphing the above function with MS Excel, we have one of the solution is w = 9

∴ (w - 9) is a factor of w³ + 42·w² + 432·w - 8019 = 0

Dividing, we get;

w² + 51·w + 891

w³ + 42·w² + 432·w - 8019 = 0

w³ - 9·w²

     51·w² - 459·w

                 891·w

                 891·w + 8019

                             0

Therefore,

w³ + 42·w² + 432·w - 8019 = 0

w³ + 42·w² + 432·w - 8019  = (x - 9)·(w² + 51·w + 891) = 0

The determinant of the factor, w² + 51·w + 891 = 51² - 4×1×891 = -963, therefore, the it has complex roots

Therefore, the real solution of (x - 9)·(w² + 51·w + 891) = 0 is w = 9

The width of the computer compartment, w = 9 inches

The length of the computer compartment, l = 24 inches + 9 inches = 33 inches

The height of the computer compartment, h = 18 inches + 9 inches = 27 inches.

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