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earnstyle [38]
3 years ago
12

A box contains 48 snack size bags of popcorn

Mathematics
1 answer:
Vika [28.1K]3 years ago
5 0

The answer should be 15/24 of an ouce beacause 35 minus 5 ounces because the box weighs 5 ounces when it is empty. Then, you need to do 30 dibided by to to get 30/48 which is reduced to 15/24.

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Joe will build a rectangular pen for his dog. A wall will form one side of the pen. Joe has 40 ft of fencing to form the other t
Svetlanka [38]
Material=wall+2*side and material is 40 ft so:

40=w+2s

w=40-2s

Area=ws, using w from above we get:

A=(40-2s)s

A=40s-2s^2

dA/ds=40-4s and d2A/ds2=-4

Since d2A/ds2 is a constant negative acceleration, when dA/ds=0, A(s) is at an absolute maximum.

dA/ds=0 when 4s=40, s=10 ft

And since w=40-2s, w=20 ft

So the dimensions of the pen are 20 ft by 10 ft, with the 20 ft side being opposite the wall.  And the maximum possible area is thus 200 ft^2
6 0
3 years ago
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A 12-ft ladder is placed 5 feet from a building. Approximately how high does the ladder reach?
ElenaW [278]
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4 0
3 years ago
How was life different for Romans with low incomes and wealthy Romans? Romans living in poverty had few duties and more free tim
ahrayia [7]

Wealthy Romans relied on servants to run their households.

brainly.com/question/16395373

8 0
2 years ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
3 years ago
3x+5=4 two step equations
s344n2d4d5 [400]

Answer:

x = -1/3

Step-by-step explanation:

Move 5 to the other side of the equation

3x = 4-5

3x = -1

x = -1/3

7 0
3 years ago
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