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mariarad [96]
3 years ago
9

Find the amplitude and the equation of the midline of the periodic function.

Mathematics
2 answers:
djverab [1.8K]3 years ago
8 0

Check the picture below.

Masja [62]3 years ago
6 0

Answer:

Option B. Amplitude =3 midline is y =2.

Step-by-step explanation:

In the graph attached we have to find the amplitude and midline of the periodic function.

Amplitude of the periodic function = (Distance between two extreme points on y asxis)/2

=  (5-(-1))/2 = (5+1)/2 =6/2 =3.

Since amplitude of this function is 3 and by definition amplitude of any periodic function is the distance between the midline and the extreme point of wave on one side.

Therefore midline of the wave function is y=2 from which measurement of the amplitude is 3.

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Hi again here's the picture
pshichka [43]

Answer:

<h2>351.88 m</h2>

Step-by-step explanation:

Given,

D = diameter of semicircular part = 42 m

L = Length of straight part = 110 m

Now, let's find the perimeter:

= 2 ( length of semi-circle + length of straight part )

= 2( \frac{\pi \: d}{2}  + l)

Plug the values

= 2( \frac{3.14  \times 42}{2}  + 110)

Calculate the product

= 2( \frac{131.88}{2}  + 110)

Divide

= 2 (65.94 + 110)

Calculate the sum

= 2 \times 175.94

Calculate

= 351.88 \: m

Hope this helps..

Best regards!!

8 0
3 years ago
Read 2 more answers
What is the value of p?
sveta [45]

Answer:

Not C, Not A, most likely B

Step-by-step explanation:

Its less than 90 because its and acute, meaning its smaller, so you can already knock off C.

6 0
3 years ago
Read 2 more answers
find a number such that if 8 is subtracted from 9 times the number , the result is 6 more than twice the number​
zheka24 [161]

Answer:

x=2

Step-by-step explanation:

let the number be x

9x-8=6+2x

9x-2x=6+8

7x=14

x=14/7

x=2

8 0
3 years ago
Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
zavuch27 [327]
Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }&#10;\\\\&#10;\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ( h, k)\\&#10;vertices\ ( h,  k\pm a)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{ a ^2 + b ^2}\\&#10;asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)&#10;\end{cases}\\\\&#10;-------------------------------

\bf \begin{cases}&#10;h=2\\&#10;k=-3\\&#10;a=6\\&#10;c=2\sqrt{10}&#10;\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2&#10;\\\\\\&#10;4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2&#10;\\\\\\&#10;\sqrt{4}=b\implies 2=b\\\\&#10;-------------------------------\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
3 years ago
Value of x-y^x-y when x=2 and y=-2?​
Nonamiya [84]

Step-by-step explanation:

x-y^x-y

2-(-2)^2-(-2)

2-(-2)^2+2

2-(-2)^4

2-(16)

=-14

3 0
3 years ago
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