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4 years ago

What numbers have an absolute value of 2 1/2

1 answer:

7 0

The absolute value of a number is that number's distance from zero. It will always be positive, even if the original number was negative.

| x | = x
| -x | = x

So,
|2 1/2| = 2 1/2
and
|-2 1/2| = 2 1/2

Hope this helps! :)

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Can I Please have some help with this question. 30 Points!!!

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Answer:

x=14

Step-by-step explanation:

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4 years ago

(02.01 LC) Solve:negative 1 over 2 x + 2 = −x + 7 (5 points)??.?.??????

jonny [76]

Answer:

x=10

Step-by-step explanation:

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3 years ago

Read 2 more answers

Minimum Average Cost

Dahasolnce [82]

Answer:

a) $\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}$

b) $\bar{C}(x)=5.81$

Step-by-step explanation:

Given that

C = 100 + 25 x - 120 ln x ,x ≥ 1.

The average cost function given as

$\bar{C}(x)=\dfrac{C(x)}{x}$

$\bar{C}(x)=\dfrac{100 + 25 x - 120 \ln x}{x}$

$\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx{x}$

Therefore

$\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}$

To find average minimum cost

$\bar{C}(x)=\dfrac{100}{x}+25-120\dfrac{lnx}{x}$

$\dfrac{d\bar{C}(x)}{dx} = -\dfrac{100}{x^2} +0-120\times \dfrac{1-lnx}{x^2}$

$0 = -\dfrac{100}{x^2} +0- 120\times \dfrac{1-lnx}{x^2}$

100 + 120 (1-lnx) = 0

$lnx=\dfrac{220}{120}$

ln x =1.833

$x=e^{1.833}$

x=6.25

$\bar{C}(x)=\dfrac{100}{6.25}+25-120\dfrac{ln6.25}{6.25}$

$\bar{C}(x)=5.81$

6 0

4 years ago

Please i need help please help me out!

TiliK225 [7]

Answer:

1. 13 or -13

2. -5 < y < -3

3. 6 or -6

4. 1/8 or -1/8

Step-by-step explanation:

Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.

The Absolute Value term is |x|

For the Negative case we'll use -(x)

For the Positive case we'll use (x)

Step 3 :

Solve the Negative Case

-(x) = 13

Multiply

-x = 13

Multiply both sides by (-1)

x = -13

Which is the solution for the Negative Case

Step 4 :

Solve the Positive Case

(x) = 13

Which is the solution for the Positive Case

Step 5 :

Wrap up the solution

x=-13

x=13

But for the case of question (2) its a different pattern..

Since this is a "less than" absolute-value inequality, my first step is to clear the absolute value according to the "less than" pattern. Then I'll solve the linear inequality.

| y + 4 | < 1

–1 < y + 4 < 1

This is the pattern for "less than". Continuing, I'll subtract 3 from all three "sides" of the inequality:

–1 – 4 < y + 4 - 4 < 1 – 4

–5 < y < -3

$- 5 < y < - 3$

The solution to the original absolute-value inequality, | y + 4 | < 1 , is the interval:

$- 5 < y < - 3$

7 0

3 years ago

Circle C with center at (−4, 6) and radius 2 is similar to circle D with center at (6, −2) and radius 4. Below is an incorrect i

BlackZzzverrR [31]

We are told that circle C has center (-4, 6) and a radius of 2.

We are told that circle D has center (6, -2) and a radius of 4.

If we move circle C's center ten units to the right and eight units down, the new center would be at (-4 + 10), (6 - 8) = (6, -2). So step 1 in the informal proof checks out - the centers are the same (which is the definition of concentric) and the shifts are right.

Let's look at our circles. Circle C has a radius of 2 and is inside circle D, whose radius is 4. Between Circle C and Circle D, the radii have a 1:2 ratio, as seen below:

$\frac{1}{2} = \frac{radius--circle C}{radius--circle D}$

If we dilate circle C by a factor of 2, it means we are expanding it and doubling it. Our circle has that 1:2 ratio, and doubling both sides gives us 2:4. The second step checks out.

Translated objects (or those that you shift) can be congruent, and dilated objects are used with similarity (where you stretch and squeeze). The third step checks out.

Thus, the argument is correct and the last choice is best.

3 0

4 years ago