All categories

3 years ago

NO LINKS OR ELSE YOU'LL BE REPORTED!Only answer if you're very good at Math.

Mathematics

1 answer:

andreev551 [17]3 years ago

4 0

Answer:

$D.\ y = 3\sin(2(x + \frac{\pi}{4}))$

Step-by-step explanation:

Given

$A = 3$ --- Amplitude

$C = \frac{\pi}{4}$ --- Right phase shift

Required

Identify the function

A sinusoidal function is defined as:

$y = A\sin(B(x + C)) + D$

Substitute values for A and C

$y = 3\sin(B(x + \frac{\pi}{4})) + D$

By comparison, only option (D) has the above format where

$y = 3\sin(2(x + \frac{\pi}{4}))$

Hence (D) is correct

You might be interested in

Rishan is building a number 3091 using Place-value blocks. There are no thousands block. How many hundreds blocks can he use ins

satela [25.4K]

Given:

The number is 3091.

To find:

The number of hundreds blocks if there are no thousands block.

Solution:

The given number is 3091.

It can be written as:

$3091=3000+0+90+1$

$3091=3\times 1000+0\times 100+9\times 10+1\times 1$

$3091=3\ Thousands+0\ Hundreds+90\ Tens+1\ Ones$

There are no thousands block. So,

$3000=30\times 100$

$3000=30\ Hundreds$

Therefore, the number of hundreds blocks in 3091 is 30.

7 0

3 years ago

6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at

SOVA2 [1]

Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

$\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\$

$Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}$