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Levart [38]
3 years ago
14

A pizza company is building a rectangular solid box to be able to deliver personal pan pizzas. The pizza company wants the volum

e of the delivery box to be 280 cubic
inches. The length of the delivery box is 6 inches less than twice the width, and the height is 2 inches less than the width. Determine the width of the delivery box.
A. 7 inches
B. 5 inches
C. 6 inches
D. 8 inches
Mathematics
1 answer:
ladessa [460]3 years ago
4 0

Here, we are required to determine the width of the delivery box given the dimensions as in the question.

The <em>correct</em> answer is choice A.

The width of the delivery box is ; w = 7.

A <em>rectangular box</em> had the shape of a cuboid.

Volume of a <em>cuboid</em> = <em>Length</em> × <em>width</em> × <em>height</em>

<em>width</em> = w

<em>Length</em> = 2w - 6

<em>height</em> = w - 2

280 = (w) × (2w - 6) × (w - 2)

280 = (2w² - 6w) × (w - 2)

280 = 2w³ - 4w² - 6w² + 12w

280 = 2w³ - 10w² + 12w

2w³ - 10w² + 12w - 280 = 0

Therefore, w³ - 5w² + 6w - 140 = 0

By testing hypothesis at w = 7, i.e (w - 7) is a factor.

The expression cam be factorized as;

(w - 7)(w² +2w + 20) = 0.

Therefore, the only real solution of w is , w = 7.

Read more:

brainly.com/question/20463446

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In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates
Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

S = ut + \frac{1}{2} at^{2}

Where S is distance

u is the initial velocity

a is acceleration

and t is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

S = S₁

u = 0 m/s ( The player starts at rest)

a = 2.2 m/s²

Then,

S₁ = 0(t) + \frac{1}{2} (2.2)t^{2}

S₁ = 1.1t^{2}

t^{2} = \frac{S_{1} }{1.1}

For player 2

S = S₂

u = 0 m/s ( The player starts at rest)

a = 1.7 m/s²

Then,

S₂ = 0(t) + \frac{1}{2} (1.7)t^{2}

S₂ = 0.85t^{2}

t^{2} = \frac{S_{2} }{0.85}

Since the time spent by both players is equal, We can  write that

t^{2} = t^{2}

Then,

\frac{S_{1} }{1.1} =  \frac{S_{2} }{0.85}

0.85S_{1} = 1.1S_{2}

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

0.85S_{1} = 1.1 ( 41 -S_{1} )

0.85S_{1} = 45.1 - 1.1S_{1}

0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1}  = 23.13 m

This is the distance covered by the first player. We can then put this value into t^{2} = \frac{S_{1} }{1.1} to determine how much time elapses before the players collide.

t^{2} = \frac{S_{1} }{1.1}

t^{2} = \frac{23.13 }{1.1}

t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs

Hence, the time that elapses before the players collide is 4.59 secs

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4 years ago
What is the equation of the following line written in general form? (The y-intercept is -1.)
Vedmedyk [2.9K]
~~~~~~~~~2x- y - 1 = 0 ~~~~~~~~~~
4 0
3 years ago
What is the b in 8b=5.6
Inga [223]
B = 0.7 because 5.6 divided by 8 equals 0.7! Hope this helpz...
3 0
4 years ago
Read 2 more answers
If i only had 2819 in my account and i now have 3349 how much more deposited?
leonid [27]

Answer:

530.

Step-by-step explanation:

If you take away 2819 from 3349, you'll get 530. That's how much more was deposited.

lmk if I'm wrong

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3 years ago
19/100 x g/10 help please :0
Rudik [331]

19g/1000 would be the right answer when simplified.

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