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nevsk [136]
3 years ago
5

Ten granola bars and twelve bottles of water cost $23. Five granola bars and four bottles of water cost $10. How much do one gra

nola bar and one bottle of water cost?
Mathematics
1 answer:
Naddika [18.5K]3 years ago
8 0

Answer:

$2.15

Step-by-step explanation:

Given :

The cost of Ten granola bars and twelve bottles of water is $23.

The cost of Five granola bars and four bottles of water is $10.

To Find : The cost of one granola bar and one bottle of water

Solution :

Let x be the cost of one granola bar.

Let y be the cost of one water bottle.

So, cost of 10 granola bars = $10 x

Cost of 5 granola bars = $5 x

Cost of twelve bottles of water = $12 y

Cost of four bottles of water = $4 y

Now we know that the cost of Ten granola bars and twelve bottles of water is $23.

So, the equation becomes:

⇒10x+12y=23  ---(a)

We also know that cost of Five granola bars and four bottles of water is $10.

So, the equation becomes:

⇒5x+4y=10  ---(b)

Now we need to solve (a) and (b) to find value of x and y

So, we will use substitution method

We will substitute value of x from (a) in (b)

⇒5(\frac{23-12y}{10} )+4y=10


⇒\frac{23-12y}{2} +4y=10


⇒11.5-6y+4y=10


⇒11.5-2y=10


⇒11.5-10=2y


⇒1.5=2y


⇒\frac{1.5}{2} =y


⇒0.75 = y


So, cost of one water bottle 'y' is $0.75

Now we are supposed to find the value of x . So, put value of y in equation (a)

⇒10x+12(0.75)=23


⇒10x+9=23


⇒10x=23-9


⇒10x= 14


⇒x= \frac{14}{10}


⇒x= 1.4

So, cost of one granola bar 'x' is $1.4

So, combine cost of 1 granola bar and one water bottle is $1.4+ $0.75 = $2.15

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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

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Answer:

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Step-by-step explanation:

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