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3 years ago

HELP ME PLEASE,

Mathematics

2 answers:

jenyasd209 [6]3 years ago

6 0

Never is )4744$, $ :!83

bagirrra123 [75]3 years ago

6 0

I'm in K12 too, Those little things in the middle of the lesson do NOT need to be completed. You can skip over those. Also if you're on the grade level where you start having "discussions" those are not necessary either.

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Write a cosine function that has a midline of 2, an amplitude of 4 and a period of pi/2.

exis [7]

y = Acos(Bx) + D;

D = 4, A = 2. Now T = 2π/B = 5π/8, B = 2π/(5π/8) = 16/5

WE get y = 2cos(16/5x) + 4

4 0

2 years ago

PLEASE HELP WILL GIVE BRAINLIEST PRECALC

nexus9112 [7]

Answer:

$270^o$ and $450^o$

Step-by-step explanation:

Recall that $\frac{\sqrt{2} }{2}$ is a value of the function cosine for the special angles: $135^o$ and $225^o$ , then:

$\frac{x}{2} =135^o\\x=270^o\\or\\ \frac{x}{2} =225^o\\x=450^o\\$

4 0

3 years ago

X = ? y = ? 16 45 degrees

mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}$

For x, we will be using the Cosine Function:

$\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}$ $(16)Cosine(45^{\circ})\text{ = x}$ $(16)(\frac{1}{\sqrt[]{2}})\text{ = x}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = x}$

Therefore, x = 8√2.

For y, we will be using the Sine Function.

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}$ $\text{ (16)Sine }(45^{\circ})\text{ = y}$ $\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = y}$

Therefore, y = 8√2.