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Maksim231197 [3]
4 years ago
5

Find the amplitude and period of the graph y=4 sin πθ

Mathematics
1 answer:
Vsevolod [243]4 years ago
7 0

Answer:

The amplitude is 4 and the period is 2

Step-by-step explanation:

In the equation y = A sin (B x)  

  • A is the amplitude, where the amplitude is the height from highest to lowest points and  divide the answer by 2
  • The period is \frac{2\pi }{B} , where the period is the distance from one peak to the next peak

∵ The equation is y = 4 sin(πФ)

- Compare it with form above

∴ A = 4 and B = π

∵ A is the amplitude

∴ The amplitude is 4

∵ Period = \frac{2\pi }{B}

∴ Period = \frac{2\pi }{\pi }

∴ Period = 2

∴ The period is 2

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180 - 101 = 79
Supplementary mean that the two angle measures add up to equal 180
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Hi I need help please
jek_recluse [69]

Answer/Step-by-step explanation:

Recall: if two figures are similar, we know that the ratio of their areas = the square of the ratio of their corresponding sides

Apply this knowledge to solve the problems given.

✔️Problem 1 (4-sided polygon):

Let the missing area be x cm²

Therefore,

\frac{x}{9} = \frac{8^2}{4^2}

\frac{x}{9} = \frac{64}{16}

\frac{x}{9} = 4

Multiply both sides by 9

x = 9*4

x = 36 cm²

✔️Problem 2 (3-sided polygon):

Let the missing area be y cm²

Therefore,

\frac{y}{240} = \frac{8^2}{32^2}

\frac{y}{240} = \frac{64}{1,024}

\frac{y}{240} = \frac{1}{16}

Multiply both sides by 240

y = ¹/16 × 240

y = 15 cm²

✔️Problem 3 (5-sided polygon):

Let the missing area be z cm²

Therefore,

\frac{z}{40} = \frac{3^2}{2^2}

\frac{z}{40} = \frac{9}{4}

Multiply both sides by 40

z = \frac{9}{4} * 40

z = \frac{9}{1} * 10

z = 90 cm²

7 0
3 years ago
Please please <br> I need help ((((riiiight now)))<br> What's the right answer and explain why!!!!!
Ilia_Sergeevich [38]
First you can apply the a²-b² = (a+b)(a-b) theorem, you'll get:

(p²)² - 9² = (p²+9)(p²-9), but that's not all, you can apply it again on the last factor:

(p²+9)(p²-3²) = (p²+9)(p+3)(p-3)

The first factor cannot be simplified like that, so that's the furthest factorization. So answer (3) is the right one.
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3 years ago
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1. Which relation represents a function?
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3 years ago
Lori creates the following design for a T-shirt.
kakasveta [241]

<u>Solution-</u>

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7


Area of rectangle 1 = 8.68 sq.in

\Rightarrow FH \times HI=8.68

\Rightarrow EB \times HI=8.68  (∵ sides of the rectangle 2)

\Rightarrow 2.8 \times HI=8.68

\Rightarrow HI=3.1


Area of Triangle 1 = 6.48 sq.in

\Rightarrow \frac{1}{2}\times AE \times EG= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48  (∵ sides of the rectangle 1)

\Rightarrow EF+3.1= 5.4

\Rightarrow EF=2.3

\Rightarrow BH=2.3  (∵ sides of the rectangle 2)


BC = BH+HI+IC

\Rightarrow 11.7= 2.3+3.1+IC

\Rightarrow IC=6.3


The area of Rectangle 2,

=EB\times BH =2.8\times 2.3=6.44\ sq.in


The area of Triangle 2,

\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in


The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in


6 0
3 years ago
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