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andreev551 [17]
3 years ago
14

Manuela solved the equation 3−2|0.5x+1.5|=2 for one solution. Her work is shown below. 3−2|0.5x+1.5|=2 −2|0.5x+1.5|=−1 |0.5x+1.5

|=0.5 0.5x+1.5=0.5 0.5x=−1 x=−2 What is the other solution to the equation? x=−6 x=−4 x=2 x=4
Mathematics
2 answers:
lions [1.4K]3 years ago
7 0

Answer:

Step-by-step explanation:

We'll just work on solving both so you can see what's involved in solving an absolute value equation. Because an absolute value is a distance, we can have that distance being both to the right on the number line of the number in question or to the left. For example, from 2 on the number line, the numbers that are 5 units away are 7 and -3. Using that logic, we will simplify the equation down so we can set up the 2 basic equations needed to solve for x.

If  3-2|.5x+1.5|=2 then

-2|.5x+1.5|=-1  What you need to remember here is that you cannot distribute into a set of absolute values like you would a set of parenthesis. The -2 needs to be divided away:

|.5x+1.5|=.5

Now we can set up the 2 main equations for this which are

.5x + 1.5 = .5  and .5x + 1.5 = -.5

Knowing that an absolute value will never equal a negative number (because absolute values are distances and distances will NEVER be negative), once we remove the absolute value signs we can in fact state that the expression on the left can be equal to a negative number on the right, like in the second equation above.

Solving the first one:

.5x = -1 so

x = -2

Solving the second one:

.5x = -2 so

x = -4

suter [353]3 years ago
4 0

Answer:

x= -4

Step-by-step explanation:

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A bucket that weighs 3 lb and a rope of negligible weight are used to draw water from a well that is 90 ft deep. The bucket is f
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Answer:

a) Lim(0-inf)  Work = (36 - 0.1*xi )*dx

b) Work = integral( (36 - 0.1*xi ) ).dx

c) Work = 2835 lb-ft

Step-by-step explanation:

Given:

- The weight of the bucket W = 3 lb

- The depth of the well d = 90 ft

- Rate of pull = 2.5 ft/s

- water flow out at a rate of = 0.25 lb/s

Find:

A. Show how to approximate the required work by a Riemann sum (let x be the height in feet above the bottom of the well. Enter xi∗as xi)

B. Express the Integral

C. Evaluate the integral

Solution:

A.

- At time t the bucket is xi = 2.5*t ft above its original depth of 90 ft but now it hold only (36 - 0.25*t) lb of water at an instantaneous time t.

- In terms of distance the bucket holds:

                           ( 36 - 0.25*(xi/2.5)) = (36 - 0.1*xi )

- Moving this constant amount of water through distance dx, we have:

                            Work = (36 - 0.1*xi )*dx

B.

The integral for the work done is:

                           Work = integral( (36 - 0.1*xi ) ).dx

Where the limits are 0 < x < 90.

C.

- Evaluate the integral as follows:

                           Work = (36xi - 0.05*xi^2 )

- Evaluate limits:

                           Work = (36*90 - 0.05*90^2 )  

                            Work = 2835 lb-ft

8 0
3 years ago
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