Can someone help me with this please. ASAP!!

I don’t know how to do this please help please!!!

Irina18 [472]

Answer:

study more so you can understand

3 0

3 years ago

How do I use the following cosine equation to get the Sinusoid Max & Min Times (x values), and Sinusoid Max and Min Values (

Paraphin [41]

9514 1404 393

Answer:

maximum: (x, y) = (12.4n+3.2, 186.322)
minimum: (x, y) = (12.4n+9.4, 163.498)

Step-by-step explanation:

You know that cos(α) is a maximum at α=0, 2π, 4π, and all even multiples of π. You know cos(α) is a minimum for α=π, 3π, 5π, and all odd multiples of π.

You can find your value of x at which y will be a maximum by setting the argument of the cosine function equal to zero (and/or 2nπ). If we use α=2nπ, then we have ...

α = (5π/31)(x -3.2) = 2nπ

(x -3.2) = (31/5)(2n) = 12.4n

Tidal maxima will occur at ...

x = 12.4n +3.2 . . . . . for integer values of n

Without bothering to go through the solution for α being odd multiples of π, we can see from this that the period is 12.4 hours. We know the tidal minimum will be half a period later, or 6.2 hours later than this.

Tidal minima will occur at ...

x = 12.4n +9.4 . . . . for integers n

__

Of course, cos(α) has extremes of ±1, so your tidal maximum will be ...

y = 11.412 +174.91 = 186.322

and your tidal minimum will be ...

y = -11.412 +174.91 = 163.498

4 0

3 years ago

At the north campus of a performing arts school, 10% of the

Alecsey [184]

There were 600 students at the north campus and 400 students at the south campus.

<h2 /><h3><u>Calculus</u></h3>

To determine, if 42% of the 1000 students at the east campus are music majors, how many students did the north and south campuses have before the merger, the following calculation must be performed:

0.10X + 0.90Y = 420
1000 x 0.42 = 420
500 x 0.10 + 500 x 0.90 = 50 + 450 = 500
600 x 0.10 + 400 x 0.90 = 60 + 360 = 420

Therefore, there were 600 students at the north campus and 400 students at the south campus.

Learn more about calculus in brainly.com/question/6581270

7 0

2 years ago

Sakura speaks 150150150 words per minute on average in hungarian, and 190190190 words per minute on average in polish. she once

son4ous [18]

The correct question is
<span>Sakura speaks 150 words per minute on average in hungarian, and 190 words per minute on average in polish. she once gave cooking instructions in hungarian, followed by cleaning instructions in polish. sakura spent 5 minutes total giving both instructions, and spoke 270 more words in polish than in hungarian. how long did sakura speak in hungarian, and how long did she speak in polish?</span>

Let
x------> total words spoken by sakura in hungarian------> 150 words /minute
y------> total words spoken by sakura in polish-----------> 190 words /minute

we know that
(x/150)+(y/190)=5--------- > equation 1
y=270 +x-------------------- > equation 2
<span>substituting 2 in 1
(x/150)+(270+x)/190=5

</span><span>multiplying all the expression by (150)*(190)
</span>190x+150*(270+x)=5*190*150
190x+40500+150x=142500
340x=102000-------------- > x=300
x=300 ------------- > total words spoken by sakura in hungarian
y=270+x=270+300=570
y=570 ----------- > total words spoken by sakura in polish

the question is <span>how long did sakura speak in hungarian, and how long did she speak in polish?
</span>
y=570 words in polish-------------------> 190 words /minute

if 190 words-----------------------------> 1 minute
570 words-------------------------- X
X=570/190=3 minutes
In polish Sakura spoke 3 minutes

x=300 words in hungarian-------------------> 150 words /minute

if 150 words-----------------------------> 1 minute
300 words-------------------------- X
X=300/150=2 minutes

In hungarian Sakura spoke 2 minutes

8 0

3 years ago

part A. logan and his friends love pepperoni pizza. Their parent's bought them one whole pizza to share .logan ate 2/6 of the pi

Vikki [24]

7/22 you add the denominators toghether and add the numerators.

5 0

3 years ago