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Licemer1 [7]
3 years ago
9

WILL GIVE 30 POINTS

Mathematics
1 answer:
Leto [7]3 years ago
7 0

Answer:

i have no idEa

Step-by-step explanation:

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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
Square root of 75 plus square root of 3
olya-2409 [2.1K]
Answer:
6√3

Explanation:
Before we begin, remember the following:
\sqrt{a*b} = \sqrt{a}  *  \sqrt{b}
m \sqrt{a} + n \sqrt{a}  = (m+n) \sqrt{a}

Now, for the given:
75 can be written as 25*3
This means that:
\sqrt{75} = \sqrt{25*3}
Applying the above concept, we can find that:
\sqrt{75} = \sqrt{25*3} = \sqrt{25}  *  \sqrt{3}
Now, we know that:
√25 = 5 (we ignored the negative value)

This means that:
√75 = 5√3

Finally, we can compute the needed sum as follows:
√75 + √3 = 5√3 + √3 = 6√3

Hope this helps :)

8 0
3 years ago
Read 2 more answers
Which statement about 6x2 + 7x – 10 is true?
amm1812

Answer:

(x+2)

Step-by-step explanation:

factor out the equation and get

(x+2)(6x-5)

7 0
3 years ago
Read 2 more answers
Which are true statements about a translation?
jeka94

a} The image is congruent to the pre-image.

c} The image could be moved left or right.

d} The image could be moved up or down.

6 0
2 years ago
If <img src="https://tex.z-dn.net/?f=tan%20%28x%29%20%3D%20%5Cfrac%7B5%7D%7B12%7D" id="TexFormula1" title="tan (x) = \frac{5}{12
Alekssandra [29.7K]

Explanation:

First, we need to find the values of the sine and cosine of x knowing the value of tan x and x being in the 3rd quadrant. Since tan x = 5/12, using Pythagorean theorem, we know that

\sin x = -\frac{5}{13}\;\;\text{and}\;\;\cos x = -\frac{12}{13}

Note that both sine and cosine are negative because x is in the 3rd quadrant.

Recall the addition identities listed below:

\sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta

\Rightarrow \sin(180+x) = \sin180\sin x + \cos180\cos x

\;\;\;\;\;\;= -\sin x = \dfrac{5}{13}

\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta

\Rightarrow \cos(180 - x) = \cos180\cos x + \sin180\sin x

\;\;\;\;\;\;=-\cos x = \dfrac{12}{13}

\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}

\Rightarrow \tan(360 - x) = \dfrac{\tan 360 - \tan x}{1 + \tan 360 \tan x}

\;\;\;\;\;\;= -\tan x = -\dfrac{5}{12}

Therefore, the expression reduces to

\sin(180+x) + \tan(360-x) + \frac{1}{\cos(180-x)}

\;\;\;\;\;= \left(\dfrac{5}{13}\right) + \left(\dfrac{5}{12}\right) + \dfrac{1}{\left(\frac{12}{13}\right)}

\;\;\;\;\;= \dfrac{49}{26}

5 0
2 years ago
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