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igor_vitrenko [27]

3 years ago

7

What is 4/5 of $45.00

1 answer:

Alex73 [517]3 years ago

4 0

I did this in my head but the answer is $36 45/5 is 9 9*4 is 36. And if you add another it totals to 45. Thus the answer is $36.

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What’s the amplitude and the period of this function?

Iteru [2.4K]

The amplitude and the period of the function are 3 and 2, respectively.

<h3>What are the amplitude and the period of a sinusoidal function?</h3>

Sinusoidal functions are periodic bounded functions whose form is described below:

y = A · sin (2π · t / T + Ф) + y'

Where:

y' - Midpoint
A - Amplitude
t - Time
Ф - Phase
T - Period

The period is the horizontal distance between two consecutive peaks or two consecutive bottoms and the amplitude is equal to the half of the distance between a peak and a consecutive bottom. Hence, the amplitude and the period are, respectively:

Amplitude

A = 0.5 · [2 - (- 4)]

A = 3

Period

T = 2 · [0.75 - (- 0.25)]

T = 2

To learn more on sinusoidal functions: brainly.com/question/24336803

#SPJ1

6 0

2 years ago

Can I get some help with this?

4vir4ik [10]

Hey !
The blank is 2- $\fbox{4}$
The blank is 6- $\fbox{3}$
The blank is 0- $\fbox{2}$

8 0

2 years ago

Read 2 more answers

The top and bottom margins of a poster are each $3$ cm and the side margins are each $2$ cm. If the area of printed material on

babymother [125]

Answer:

Therefore the dimension of the poster is 12 cm by 8 cm.

Step-by-step explanation:

Let the length of the poster be x and the width be y.

Given that the area of the poster is 96 cm².

∴xy =96

$\Rightarrow y= \frac{96}{x}$

The sides margins each are 2 cm and the top and bottom margins of the poster are each 3 cm.

The length of printing space is =(x- 2.3) cm

= (x-6) cm

The width of the printing space is =(y-2.2) cm

=( y-4 )cm

The area of the printing space is A=(x-6)(y-4) cm²

∴A=(x-6)(y-4)

$\Rightarrow A=(x-6)(\frac{96}{x}-4)$ [ Putting $y= \frac{96}{x}$ ]

$\Rightarrow A=120-\frac{576}{x}-4x$

Differentiating with respect to x

$A'= \frac{576}{x^2}-4$

Again differentiating with respect to x

$A''=-\frac{1152}{x^3}$

To find the minimum area, we set A'=0

$\therefore \frac{576}{x^2}-4=0$

$\Rightarrow \frac{576}{x^2}=4$

$\Rightarrow x^2=\frac{576}{4}$

$\Rightarrow x^2 =144$

$\Rightarrow x=\pm 12$

Dimension can't be negative.

Therefore x=12

If x=12, the value of A''>0,then at x=12, the area of the poster will be minimum.

If x=12, the value of A''<0,then at x=12, the area of the poster will be minimum.

$\therefore A''|_{x=12}=-\frac{1152}{12^3}$

Therefore at x= 12 cm the area of the poster will be maximum.

The width of the poster is $y=\frac{96}{12}$ = 8 cm

Therefore the dimension of the poster is 12 cm by 8 cm.

3 0

3 years ago

21. Create an area model to represent<br> 14 X 17. Find the product.

elena-s [515]

14 times 17 is 238.

I will draw an area model and add it to my answer.

Hope this helps!!!

3 0

3 years ago

Please help tysm<br><br> what are the A, B and C for the following quadratic<br><br> y=x^2-3x+10

grigory [225]

A=x^2
B=-3x
C=10
Thats the a,b, and c for that question

8 0

3 years ago