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BaLLatris [955]
3 years ago
6

Find a positive number such that the sum of the number and its reciprocal is as small as possible?

Mathematics
1 answer:
WARRIOR [948]3 years ago
5 0
You're minimizing the function f(x)=x+\dfrac1x where x>0.

Take the derivative, set to zero and find the critical points:

f'(x)=1-\dfrac1{x^2}=0\implies1=\dfrac1{x^2}

This has two roots, x=\pm1, but we ignore the negative root.

Check the signs of the derivative to the left and right of x=1 and you'll see that f'(x) for x, and f'(x)>0 for x>1, which means a minimum occurs at x=1.
You might be interested in
Evaluate the expression and enter your answer in the box below.<br> 8 + -9<br> Answer here
OLEGan [10]

Answer:

Step-by-step explanation:

remember plus minus = minus

so 8+-=9

8-9=-1

8 0
3 years ago
What are the coordinates of the focus of the parabola?
Vikki [24]

Answer:

(-8,2)

hope it help :)

8 0
3 years ago
Could I get a hand with this, please?
Mazyrski [523]

Answer:

do no i just need points

Step-by-step explanation:

3 0
3 years ago
What is the equation of a circle with center of (4,2) passing through the point (7,2)
attashe74 [19]

Answer:

(x-4)^2 + (y-2)^2 = 3^2

or

(x-4)^2 + (y-2)^2 = 9 (simplified if needed)

Step-by-step explanation:

-Equation of a circle is:

(x-h)^2+(y-k)^2=r^2 where the center is (h, k) and the radius is r^2.

-Place the center and the point onto that equation:

(7-4)^2+(2-2)^2=r^2

-Then, you solve:

(7-4)^2+(2-2)^2=r^2

(3)^2+(0)^2=r^2

9 +0=r^2

9=r^2

\sqrt{9} =\sqrt{r^2}

3=r

-So, the result is:

(x-4)^2+(y-2)^2=3^2

or

(x-4)^2+(y-2)^2=9 (simplified if needed)

3 0
3 years ago
5/6÷5=___<br><br>3/7÷6=___<br><br>5/8÷8=___<br><br>15/8÷5​=___
alexandr1967 [171]

Answer:

1/6

1/14

5/64

3/8

Step-by-step explanation:

Easiest and fastest way to evaluate these is to plug it into a calc.

7 0
3 years ago
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