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3 years ago

1.What is the frequency of the sinusoidal graph?

Mathematics

2 answers:

sineoko [7]3 years ago

6 0

Answer:

1. " second picture"

2. The graph of h(x) is shown below.

Step-by-step explanation:

The given function is

h(x)=7\sin xh(x)=7sinx

The general form of sine function is

f(x)=a\sin(bx+c)+df(x)=asin(bx+c)+d

Where, a is amplitude, b is period, c is phase shift and d is vertical shift.

So, the amplitude of the given function is 7, period is 1, phase shift is 0 and vertical shift is 0.

It means the minimum value of function is -7 and maximum value is 7.

Put x=0 in the given function.

h(x)=7\sin (0)=7(0)=0h(x)=7sin(0)=7(0)=0

Put x=-\frac{\pi}{2}x=− 2ππ in the given function.

h(x)=7\sin (-\frac{\pi}{2})=-7(1)=-7h(x)=7sin(−2π )=−7(1)=−7

Put x=\frac{\pi}{2}x= 2π in the given function.

h(x)=7\sin (\frac{\pi}{2})=7(1)=7h(x)=7sin( 2π )=7(1)=7

Therefore the points on the function are (0,0), (-\frac{\pi}{2},-7),(\frac{\pi}{2},7)(− 2π ,−7),( 2π,7) .

The graph of function is shown below.

"Hope this helps"

Elodia [21]3 years ago

6 0

Answer:

Step-by-step explanation:

maximum is at 7, minimum is at -7, midline is at pie

Read more about scale factors at:

brainly.com/question/15891755

#SPJ1

4 0

2 years ago

Fine slope intercept form <br> y = x+7 (2, 3)<br> Help please and thanks

Dimas [21]

Answer:

y=x7 X 5

Step-by-step explanation:

6 0

3 years ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .