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3 years ago

I need help.................

Mathematics

1 answer:

cluponka [151]3 years ago

4 0

Answer:

(1, 5)

Step-by-step explanation:

The midpoint of a line segment has coordinates that are the average of the end-point coordinates.

M = (X + Y)/2 = ((-3, 4) +(5, 6))/2 = (-3+5, 4+6)/2 = (2, 10)/2

M = (1, 5)

The midpoint is (1, 5).

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Answer questions 13. And 14. Please! Show minimal work...

ale4655 [162]

13. The equation of the given function is $y=4sin[\frac{(t-\frac{4}{3})}{2}]-2$

14. The equation of the cotangent function is $y=cot[2(t-\frac{1}{3})]+2$

Step-by-step explanation:

Let us revise the transformation of the trigonometric function:

y = a f[b(x + c)] + d, where

Amplitude is a
f represents the trigonometry function
Period is 2π/b
Phase shift is c (positive is to the left)
Vertical shift is d

13.

∵ y = a sin( $\frac{2\pi t}{T}$ ), where a is the amplitude

and T is the wave in seconds

∵ The amplitude is 4

∴ a = 4

∵ The period is 4π

∴ T = 4π

From the rules above

∵ The period is 2π/b

∴ T = $\frac{2\pi }{B}$

∴ 4π = $\frac{2\pi }{b}$

- By using cross multiplication

∴ 4π(b) = 2π

- Divide both sides by 4π

∴ b = $\frac{1}{2}$

∵ The phase shift is $-\frac{4}{3}\pi$

∵ c is the phase shift

∴ c = $-\frac{4}{3}\pi$

∵ The vertical shift is -2

∵ d is the vertical shift

∴ d = -2

Now substitutes the values of a, b, c and d in the form of the equation below

∵ y = a sin[b(t + c)] + d

∴ $y=4sin[\frac{1}{2}(t-\frac{4}{3})]-2$

You can write it as $y=4sin[\frac{(t-\frac{4}{3})}{2}]-2$

The equation of the given function is $y=4sin[\frac{(t-\frac{4}{3})}{2}]-2$

14.

y = cot[b(t + c)] + d

∵ The period = π

∵ The period is 2π/b

- Equate π by 2π/b to find b

∴ π = $\frac{2\pi }{b}$

- By using cross multiplication

∴ π(b) = 2π

- Divide both sides by π

∴ b = 2

∵ The phase shift is $-\frac{1}{3}\pi$

∵ c is the phase shift

∴ c = $-\frac{1}{3}\pi$

∵ The vertical shift is 2

∵ d is the vertical shift

∴ d = 2

Now substitutes the values of b, c and d in the form of the equation below

∵ y = cot[b(t + c)] + d

∴ $y=cot[2(t-\frac{1}{3})]+2$

The equation of the cotangent function is $y=cot[2(t-\frac{1}{3})]+2$

Learn more:

You can learn more about trigonometry function in brainly.com/question/3568205

#LearnwithBrainly

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