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2 years ago

What is the maximum value that the graph of y = cos x assumes?

2 answers:

8 0

The maximum value of that graph is 1 and we can prove this by means of this graph:
http://assets.openstudy.com/updates/attachments/4fe27e44e4b06e92b87169f6-syderitic-1340243754047-unt...

I hope this one works for you

iogann1982 [59]2 years ago

4 0

Answer:

maximum value that the graph y = cos x assumes is 1

Step-by-step explanation:

y =cos x is the function given.

By definition of trignometric ratios, cos x = adjacent side/hypotenuse of a right triangle.

Obviously hypotenuse can never be less than a side and hence cos x can take maximum value as 1 only

cosx is a periodic function with period 2pi and maximum value 1

maximum value that the graph of y = cos x assumes=1

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Brianna left the bookstore traveling 32 mph. Then 3.2 hours later, Sonya left traveling the same direction at 44.8 mph. How long

Lostsunrise [7]

Answer:

8 hours

Step-by-step explanation:

Let t hours be the time needed Sonya to catch up Brianna.

Sonya:

Time = t hours

Rate = 44.8 mph

Distance $=D_S$ miles

Brianna

Time $=t+3.2$ hours

Rate = 32 mph

Distance $=D_B$ miles

Note that $D_S=D_B,$ because Sonya must catch up Brianna.

Use formula $D=r\cdot t$

Hence,

$44.8t=32(t+3.2)$

Solve this equation:

$44.8t=32t+32\cdot 3.2\\ \\44.8t-32t=102.4\\ \\12.8t=102.4\\ \\t=8$

3 0

2 years ago

A small radio transmitter broadcasts in a 29 mile radius. If you drive along a straight line from a city 32 miles north of the t

hram777 [196]

Answer: He will pick up a signal from the transmitter 29.43 miles of the drive.

Step-by-step explanation:

Since we have given that

Radius = 29 mile

So, equation of circle would be

$x^2+y^2=29^2=841$

There is a straight line from a city 32 miles north i.e. (0,32) to a second city 40 miles east of the transmitter i.e. (40,0)

So, equation of line would be

$y-32=\dfrac{0-32}{40-0}(x-0)\\\\y-32=\dfrac{-32}{40}x\\\\y-32=-0.8x\\\\y+0.8x=32\\\\y=32-0.8x$

So, we will put this value in the equation of circle.

$x^2+(32-0.8x)^2=841\\\\x^2+1024+0.64x^2-51.2x=841\\\\1.64x^2-51.2x=841-1024\\\\1.64x^2-51.2x=-183\\\\1.64x^2-51.2x+183=0$

So, x= 4.12 and x = 27.1

So, y = 28.7, y = 10.32

So, distance between them would be

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\=\sqrt{(27.1-4.12)^2+(28.7-10.32)^2}\\\\=29.43$

Hence, he will pick up a signal from the transmitter 29.43 miles of the drive.

7 0

2 years ago

Timed math quiz<br> Please help me

Marianna [84]

Answer:

-3,2

Step-by-step explanation:

Its the points on x axis

5 0

2 years ago

Read 2 more answers

Use a Pythagorean identity to rewrite the equation below using only the function sin θ. Then find the value of r if θ=30°,60°, a

zepelin [54]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------\\\\
r=4sin(\theta )cos^2(\theta )\implies r=4sin(\theta )[1-sin^2(\theta )]
\\\\\\
r=4sin(\theta )-4sin^3(\theta )\qquad \qquad 
\begin{cases}
r=4sin(30^o)-4sin^3(30^o)\\
r=4sin(60^o)-4sin^3(60^o)\\
r=4sin(90^o)-4sin^3(90^o)
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$\bf \begin{cases}
r=4\left( \frac{1}{2} \right)-4\left( \frac{1}{2} \right)^3\\\\
r=4\left( \frac{\sqrt{3}}{2} \right)-4\left( \frac{\sqrt{3}}{2} \right)^3\\\\
r=4(1)-4(1)^3
\end{cases}\implies 
\begin{cases}
r=2-\frac{1}{2}\\\\
r=2\sqrt{3}-\frac{3\sqrt{3}}{2}\\\\
r=4-4
\end{cases}\implies 
\begin{cases}
r=\frac{1}{2}\\\\
r=\frac{\sqrt{3}}{2}\\\\
r=0
\end{cases}$

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2 years ago

How to set up a linear equation from a chart

balandron [24]

Start anywhere basically, just keep the line to where it will always continue at a constant value. Dont curve the line for example, that would make it unlinear.

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2 years ago