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Sav [38]
3 years ago
6

How do I solve this?

Mathematics
1 answer:
andreev551 [17]3 years ago
4 0

Answer:

x is in the range [-1,4]

Step-by-step explanation:

I haven't worked with absolute value inequalities in awhile, but let's take a wack at this.

We are given the following inequality:

| 2x - 3 | <= 5

This implies two possible cases:

[1] -5 <= 2x -3

Or

[2] 2x - 3 <= 5

So let's solve x for both of these cases:

[1] -5 <= 2x - 3

 -2 <= 2x

 -1  <= x

[2]  2x - 3 <= 5

 2x <= 8

  x <= 4

So from these cases, we can say the following is true:

x >= -1  and x <= 4

Thus, we can write this in the form

-1 <= x <= 4

Or in interval notation:

{ x is element of reals | -1 <= x <= 4}

Also written as

x is in the range [-1,4]

Where the closed brackets represent 1 and 4 as possible answers whereas parenthesis would imply they were not.

Cheers.

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4 years ago
f the price charged for a candy bar is​ p(x) cents, where p (x )equals 162 minus StartFraction x Over 10 EndFraction ​, then x t
andreev551 [17]

Answer:

a. 1620-x^2

b. x=810

c. Maximum value revenue=$656,100

Step-by-step explanation:

(a) Total revenue from sale of x thousand candy bars

P(x)=162 - x/10

Price of a candy bar=p(x)/100 in dollars

1000 candy bars will be sold for

=1000×p(x)/100

=10*p(x)

x thousand candy bars will be

Revenue=price × quantity

=10p(x)*x

=10(162-x/10) * x

=10( 1620-x/10) * x

=1620-x * x

=1620x-x^2

R(x)=1620x-x^2

(b) Value of x that leads to maximum revenue

R(x)=1620x-x^2

R'(x)=1620-2x

If R'(x)=0

Then,

1620-2x=0

1620=2x

Divide both sides by 2

810=x

x=810

(C) find the maximum revenue

R(x)=1620x-x^2

R(810)=1620x-x^2

=1620(810)-810^2

=1,312,200-656,100

=$656,100

7 0
4 years ago
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