= 3 - 3cos x - 2(1- cos^2 x)
= 2cos^2 x - 3cos x + 1
=(cos x -1)(cosx - 1/2)
now cos x = 1 or cos x= 1/2
draw a graph we get = + ve values
hence .. 0 ,180,360 and 60,240, degrees in range x<2pi

6 0

4 years ago

Which is equivalent to the following expression? (2x^2+4xy-y^2)+(x^2+3xy-y^2)

Karo-lina-s [1.5K]

Answer C : 3x^2+7xy-2y^2

equivalent to the following expression

$(2x^2+4xy-y^2)+(x^2+3xy-y^2)$

$2x^2+4xy-y^2+x^2+3xy-y^2$

Combine like terms

$2x^2+x^2+4xy+3xy-y^2-y^2$

2x^2 + x^2 is 3x^2

4xy+3xy is 7xy

-y^2-y^2 is -2y^2

So expression becomes

$3x^2+7xy-2y^2$

Answer C is correct

3 0

4 years ago

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At which value in the domain does f(x) = 0?

Marrrta [24]

Answer:-infinity,infinity

Step-by-step explanation:

None

3 0

3 years ago

Find the value of x.<br> please help me!!!!!!!!

Zanzabum

Answer:

x = 55° (nearest degree)

Step-by-step explanation:

Trig ratios:

$sin(\theta)=\dfrac{O}{H} \ \ \ cos(\theta)=\dfrac{A}{H} \ \ \ tan(\theta)=\dfrac{O}{A} \ \ \$

where $\theta$ is the angle, O is the side opposite the angle, A is the side adjacent to the angle, and H is the hypotenuse of a right triangle

Given:

$\theta=x$
A = 3.3
H = 5.7

$\implies cos(x)=\dfrac{3.3}{5.7}$

$\implies x=\arccos \dfrac{3.3}{5.7}=54.623459...$

⇒ x = 55° (nearest degree)

5 0

2 years ago

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