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3 years ago

Solve the equation on the interval [0, 2pi): 12cos^2 x-9=0 What to do?

Mathematics

1 answer:

siniylev [52]3 years ago

6 0

Answer:

$\large\boxed{x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}\ \vee\ x=\dfrac{7\pi}{6}\ \vee\ x=\dfrac{11\pi}{6}}$

Step-by-step explanation:

$12\cos^2x-9=0\qquad\text{add 9 to both sides}\\\\12\cos^2x=9\qquad\text{divide both sides by 12}\\\\\cos^2x=\dfrac{9}{12}\\\\\cos^2x=\dfrac{9:3}{12:3}\\\\\cos^2x=\dfrac{3}{4}\to\cos x=\pm\sqrt{\dfrac{3}{4}}\\\\\cos x=\pm\dfrac{\sqrt3}{2}\\\\\text{Look at the picture}\\\\x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}\ \vee\ x=\dfrac{7\pi}{6}\ \vee\ x=\dfrac{11\pi}{6}$

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eimsori [14]

Answer:

I think its D

Step-by-step explanation:

You can rule out A and B because he would be able to throw it farther than 30 ft do that leaves C and D but davis Couldn't throw a 9.1 lb ball 30 ft so the answer is D

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3 years ago

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

velikii [3]

Answer:

The maximum value= 36

Minimum value = - 36

Step-by-step explanation:

Given that

f(x, y, z) = 8 x + 8 y + 4 z

h(x,y,z)=4 x² + 4 y² + 4 z² - 36

From Lagrange multipliers

Δf = λ Δh

Δf = < 8 ,8 , 4>

Δh = < 8 x ,8 y , 8 z>

Δf = λ Δh

< 8 ,8 , 4> = < 8 λ x ,8 λ y , 8 λ z>

8 = 8 λ x -------------1

8 = 8 λ y ---- ------2

4 = 8 λ z ----------------3

From equation 1 ,2 and 3

Now by putting the value of x,y and z in the following equation

4 x² + 4 y² + 4 z² = 36

$4\times \dfrac{1}{\lambda^2 }+4\times \dfrac{1}{\lambda^2 }+4\times \dfrac{1}{(2\lambda)^2 }=36$

$\dfrac{4}{\lambda^2 }+ \dfrac{4}{\lambda^2 }+ \dfrac{1}{\lambda^2 }=36$

So the value of λ is

$\lambda =\pm \dfrac{1}{2}$

When λ = 1/2

x = 1 / λ , y=1 / λ , z= 1 /2 λ

x= 2 , y = 2 , z=1

f(x, y, z) = 8 x + 8 y + 4 z

f(2, 2, 1) = 8 x 2 + 8 x 2 + 4 x 1

f(2, 2, 1) =36

When λ = - 1/2

x = 1 / λ , y=1 / λ , z= 1 /2 λ

x= - 2 , y = - 2 , z= - 1

f(x, y, z) = 8 x + 8 y + 4 z

f(-2, -2, -1) = 8 x (-2) + 8 x (-2) + 4 x (-1)

f(-2, -2, -1) = - 36

The maximum value= 36

Minimum value = - 36

7 0

3 years ago

Read 2 more answers

Awnser photo quick please #3

eduard

Answer:

Step-by-step explanation:

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3 years ago

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Doss [256]

Answer:

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Then we need to solve:

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x = √(-3)

This is the square root of a negative number, then this is a complex number.

This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.

D: x ∈ R.

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r(x) = 1/4

Then we need to solve:

$\frac{1}{4} = \frac{x + 1}{x^2 + 3}$