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enot [183]
3 years ago
15

On a number line, suppose the coordinate of A is 0 and AR=3. What are possible coordinates of the midpoint of AR?

Mathematics
1 answer:
Vika [28.1K]3 years ago
6 0

Answer:

1.5 and -1.5

Step-by-step explanation:

The coordinate of A is 0. It is given that AR = 3. AR represents the distance from point A to point R. Remember that distance is always positive. So, from here we have two possibilities.

  • Point R is on Right side of A i.e. at coordinate 3
  • Point R is on Left side of A i.e. at coordinate -3

In both of these cases AR which is the distance from A to R will be 3.

So, if R is on Right side of A, the midpoint of AR would be:

\frac{0+3}{2}=\frac{3}{2}=1.5

Or if R is on Left side of A, the midpoint of AR would be:

\frac{0+(-3)}{2}=\frac{0-3}{2}=-\frac{3}{2}=-1.5

Thus, the possible coordinates of the midpoint of AR are 1.5 and -1.5

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Let total amount spent =x

Dita spent 35 % of total amount.

35% of x = 0.35x

So amount spent by Dita = 0.35x

% of money spent by Benton = 20 % of x = 0.2x

We are given that Benton spent $12 more than Claire,

so making equation we have ,

Money spent by Dita = Money spent by Claire +12

0.35x =0.2x +12

0.35x -0.2x = 12

0.5x =12

x=80

So total amount spent was $80.

Money spent by Benton = 0.2x = 0.2 * 80= $16

So amount spent by Benton is $16.



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Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
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Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

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