I need to know how to answer theses type of problems!

IgorLugansk [536]

The graph of $\mathbf{y=\sin{3x}}$ is given by,

Here given the function is $y=\sin{3x}$

when $x=0\Rightarrow y=\sin{3\times 0}=\sin0=0$ , then it passes through the origin (0,0).

When $x=\frac{\pi}{6}\Rightarrow y=\sin{3\times\frac{\pi}{6}}=\sin\frac{\pi}{2}=1$

Since we know that, $-1\leq\sin{x}\leq1,\forall x$ , then the function has maximum height at $x=\frac{\pi}{6}$ .

Again when $x=\frac{\pi}{3}\Rightarrow y=\sin{(3\times\frac{\pi}{3})}=\sin\pi=0$ , again the function is 0.

So clearly the function is the Oscillating function.

Using graphing calculator we get the graph of function $y=\sin3x$ ,

Learn more about Trigonometric Function here -

brainly.com/question/1143565

#SPJ10

8 0

2 years ago

Which expressions are equivalent to 4(4x + 2y + x − y)? (1 point)

Citrus2011 [14]

=4(4x+2y+x-y)
=4*4x+4*2y+4*x-4*y
=16x+8y+4x-4y
=16x+4x+8y-4y
=20x+4y

donc la bonne reponse est la 3
4(5x+y)*
=4*5x+4*y
=20x+4y

5 0

3 years ago

How to solve the equation 8b - 12 = 5b

Virty [35]

Answer:

b = 4

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Equality Properties

Step-by-step explanation:

Step 1: Define equation

8b - 12 = 5b

Step 2: Solve for b

Subtract 5b on both sides: 3b - 12 = 0
Add 12 on both sides: 3b = 12
Divide 3 on both sides: b = 4

Step 3: Check

Plug in b to verify it's a solution.

Substitute: 8(4) - 12 = 5(4)
Multiply: 32 - 12 = 20
Subtract: 20 = 20

And we have our final answer!

4 0

3 years ago

Use series to verify that <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Which equation represents a line that passes through the two points in the table?

jekas [21]

Answer:

C. y - 5= 5/3 ( x - 3 )

Step-by-step explanation:

I used a graphing calculator

5 0

3 years ago