Help me in question 6
1 answer:
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Answer:
Cost of a snack size bag of chips = $0.8
Cost of a water bottle = $1.6
Cost of a gallon of milk = $3.4
Step-by-step explanation:
Let the cost of water bottle = $a
Cost of 1 gallon milk = $b
and cost of 1 bag of chips = $c
So, cost of 4 gallons of milk, 6 bottles of water and 5 bags of chips = $(4b + 6a + 5c)
4b + 6a + 5c = 27.2 -----(1)
Cost of one water bottle = 2 × cost of one bag of chips
a = 2c -----(2)
A gallon of milk = cost of water bottle + $1.80
b = a + 1.8
From equation (2)
b = 2c + 1.8 -----(3)
Now we substitute these values from equations (2) and (3) in equation (1)
4(2c + 1.8) + 6(2c) + 5c = 27.2
8c + 7.2 + 12c + 5c = 27.2
25c = 27.2 - 7.2
c =
c = $0.8
From equation (2)
a = 2×0.8
a = $1.6
From equation (3)
b = 2(0.8) + 1.8
b = 1.6 + 1.8
b = $3.4
Therefore, cost of a snack size bag of chips = $0.8
Cost of a water bottle = $1.6
Cost of a gallon of milk = $3.4
Define
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s