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3 years ago

12

|2w - 3| > 5 This is solving absolute value Inequalities. you through step by step?

1 answer:

dusya [7]3 years ago

5 0

If |2w - 3| > 5

then

2w - 3 < - 5 or 2w - 3 > 5

2w < - 5 + 3 or 2w > 5 + 3

2w < - 2 or 2w > 8

w < -2/2 or w > 8/2

w < -1 or w > 4

So the solution set is

S = { x E R : w < -1 or w > 4}

or using the interval notation

S = (- infinity, -1) U (4, + infinity).

I hope this helps. =)

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Step-by-step explanation:

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Read 2 more answers

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

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

Use $x=1.5 t$ and $F=42-0.15 t$

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

$=\int_{0}^{36} 63-0.225 t d t$

$=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}$

$=\left[63 t-0.2 t^{2}\right]_{0}^{36}$

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

4 0

3 years ago