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2 years ago

Which of the following cosine functions has a period of 3 π ?

Mathematics

2 answers:

Alinara [238K]2 years ago

7 0

Answer:

c. y= cos2/3x

Step-by-step explanation:

grin007 [14]2 years ago

5 0

ANSWER

$y = \: \cos( \frac{2}{3}x )$

EXPLANATION

The period of a cosine function is calculated using the formula;

$T= \frac{2 \pi}{b}$

Therefore , if the period is 3π, then

$\frac{2 \pi}{b} = 3\pi$

We solve b to obtain,

$b = \frac{2}{3}$

This implies that, the given cosine function must have an equation of the form

$y = a \: \cos( \frac{2}{3}x )$

Among the options, the function in this form is:

$y = \: \cos( \frac{2}{3}x )$

The correct choice is C.

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Find the roots of the polynomial equation x^3+2x^2+3x+6=0

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Answer: x=−2

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4 0

2 years ago

Find the solutions to x⁴-5x²-36=0 and the x-intercepts of the graph of y=x⁴-5x²-36.

disa [49]

Answer:

The solutions $x^4-5x^2-36=0$ are $x=3,\:x=-3,\:x=2i,\:x=-2i$ and the x-intercepts of $y=x^4-5x^2-36$ are $x=3,\:x=-3$

Step-by-step explanation:

Finding the solutions to $x^4-5x^2-36$ means finding the roots, a root is where the function is equal to zero.

The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero.

To find the roots you need to:

Rewrite the equation with $u=x^2$ and $u^2=x^4$

$u^2-5u-36=0$

Solve by factoring

$\mathrm{Break\:the\:expression\:into\:groups}\\u^2-5u-36=\left(u^2+4u\right)+\left(-9u-36\right)$

$\mathrm{Factor\:out\:}u\mathrm{\:from\:}u^2+4u=u\left(u+4\right)$

$\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9u-36=-9\left(u+4\right)$

$u^2-5u-36=u\left(u+4\right)-9\left(u+4\right)$

$\mathrm{Factor\:out\:common\:term\:}u+4\\\left(u+4\right)\left(u-9\right)$

$u^2-5u-36=\left(u+4\right)\left(u-9\right)=0$

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions to the quadratic equation are:

$\:u=-4,\:u=9$

Substitute back $u=x^2$ , solve for x

$x^4-5x^2-36=(u-9)(u+4)=(x^2-9)(x^2+4)$

Apply the difference of squares formula

$x^4-5x^2-36=(x^2-9)(x^2+4)=(x-3)(x+3)(x^2+4)$

$(x-3)(x+3)(x^2+4)=0$

Using the Zero factor Theorem: if ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0)

The solutions are:

$\:x=3,\:x=-3,\:x=2i,\:x=-2i$

Because two of the solutions are complex roots the only x-intercepts are $x=3,\:x=-3$

3 0

3 years ago