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3 years ago

What is the minimum value of the function g (2) = x2 - 6x – 12? -21 3+ V21 3 3-21

Mathematics

2 answers:

kirill [66]3 years ago

7 0

Answer:

g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12? g (2) = x2 - 6x – 12?

Step-by-step explanation:

zubka84 [21]3 years ago

7 0

Answer:

minimum value is -21; ordered pair of min value is (3, -21)

Step-by-step explanation:

graph the function and find the lowest point on the curve

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3 years ago

The degree of 5x4y2z3 is 9.<br> O True<br> O False

pashok25 [27]

Answer:

true

Step-by-step explanation:

because the expression shown is a monomial we simply add the degrees of each term

4 + 2 + 3 = 9

5x4y2z3 indeed has a degree of 9

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3 years ago

Read 2 more answers

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

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2 years ago