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sp2606 [1]
3 years ago
6

Mr. Rose spent $63 for a sport jacket and a pair of slacks. If the jacket cost $33 more than the slacks, how much did he pay for

each?
Which system of equations represents the word problem if j is the jacket price and s is the price of the slacks?

j + s = 63 and s - j = 33
j + s = 63 and j - s = 33
js = 63 andj/s = 33
Mathematics
2 answers:
julia-pushkina [17]3 years ago
8 0

Answer:

Option B is correct

j+s = 63

j-s = 33

Step-by-step explanation:

Here, j represents the jacket price and s represents the slacks price.

As per the statement:

Mr. Rose spent $63 for a sport jacket and a pair of slacks.

⇒j+s = 63

It is also given that:

If the jacket cost $33 more than the slacks

⇒ j = 33 +s

or

j-s = 33      

Then the system of equations is:

j+s = 63

j-s = 33

which represents the given word problem

alexdok [17]3 years ago
3 0
J + s = 63 and j - s = 33 because if the jacket cost 33 more than the slacks then you start with the jacket and subtract the slacks which are 33 less than the jacket you have 33.  


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74.88 x 8 rounded to the nearest cent
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Answer:

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Step-by-step explanation:

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A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f
zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time t

At t=0, the weight of the bucket is 6+36=42 \mathrm{lb}

After t seconds, the weight of the bucket is 42-0.15 t \mathrm{lb}

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after t seconds is x=1.5 t

Since the well is 80 ft deep, the time to pull out the bucket is \frac{80}{2}=40 \mathrm{~s}

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

W=\int_{t=0}^{t=36} F d x

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W=\int_{0}^{36}(42-0.15 t)(1.5 d t)

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W=3200 \mathrm{ft} \cdot \mathrm{lb}

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