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Alex_Xolod [135]
3 years ago
7

Simplify the radical expression.

Mathematics
1 answer:
levacccp [35]3 years ago
6 0

Answer:

9√3

Step-by-step explanation:

√(81*3)

√81*√3

9√3

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Find the minimum and maximum value of the function on the given interval by comparing values at the critical points and endpoint
Kitty [74]

Answer:

maximum: y = 1

minimum: y = 0.

Step-by-step explanation:

Here we have the function:

y = f(x) =  √(1 + x^2 - 2x)

we want to find the minimum and maximum in the segment [0, 1]

First, we evaluate in the endpoints, which are 0 and 1.

f(0)  =√(1 + 0^2 - 2*0) = 1

f(1) = √(1 + 1^2 - 2*1) = 0

Now let's look at the critical points (the zeros of the first derivate)

To derivate our function, we can use the chain rule:

f(x) = h(g(x))

then

f'(x) = h'(g(x))*g(x)

Here we can define:

h(x) = √x

g(x) = 1 + x^2 - 2x

Then:

f(x) = h(g(x))

f'(x)  =  1/2*( 1 + x^2 - 2x)*(2x - 2)

f'(x) = (1 + x^2 - 2x)*(x - 1)

f'(x) = x^3 - 3x^2 + x - 1

this function does not have any zero in the segment [0, 1] (you can look it in the image below)

Thus, the function does not have critical points in the segment.

Then the maximum and minimum are given by the endpoints.

The maximum is 1 (when x = 0)

the minimum is 0 (when x = 1)

7 0
3 years ago
1/2(4y + 18) + 2 simplified
Vanyuwa [196]

Answer: 2y+11

Step-by-step explanation:

I hope this helps<3

3 0
3 years ago
Can someone explain to me how to do this and give me the answer?
storchak [24]

x + y < = 200
1500x + 1250y < = 175,000
7 0
3 years ago
Help plz due today and don't worry abt my answers
Elanso [62]
This is for the one with the green background :) I corrected a few of the errors that collided with the answers for the 3 qns you didn’t know. Hope this helps!!!

8 0
3 years ago
using the square root property and the principle root, we can take the square root of both sides to get
Alchen [17]
<span><span><span>x = ± 2</span></span><span><span>Then the solution is </span><span>x = ± 2</span></span><span><span>hope this helps</span></span></span>
4 0
3 years ago
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