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lara31 [8.8K]
2 years ago
7

What is the amplitude of the sinusoidal function?

Mathematics
1 answer:
dem82 [27]2 years ago
7 0

Answer:

The amplitude of the sinusoidal function is 3

Step-by-step explanation:

Amplitude=\frac{Maximum+Minimum}{2}\\\\Amplitude=\frac{1+(-7)}{2}\\ \\Amplitude=\frac{1-7}{2}\\\\Amplitude=\frac{-6}{2}\\ \\Amplitude=-3=3

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A package delivery service divides their packages into weight classes. suppose that packages in the 14 to 20 pound class are uni
nexus9112 [7]

Answer: yy - \alpha \alpha = Percentile (\beta \beta -\alpha \alpha )

yy = 0.40 (20 - 14) + 14[tex] yy = 2.4 + 14

yy = 16.4

Thus, 16.4 pound class belongs to the 40th percentile of the weight class.

6 0
3 years ago
I need an answer quickly!
alexandr1967 [171]

Answer:

d=2√13

Step-by-step explanation:

B(1,4) T(4,6)  what is BD

find BT

d=√(x2-x1)²+(y2-y1)²

d=√(4-1)²+(6-4)²

d=√3²+2²

d=√9+4

d=√13 this is haf the distance

BD=2d=2√13

6 0
3 years ago
Ivan draws PQR on the coordinate plane.
Ivahew [28]

Answer:

Perimeter of PQR = 37 units (Approx.)

Step-by-step explanation:

Using graph;

Coordinate of P = (-2 , -4)

Coordinate of Q = (16 , -4)

Coordinate of R = (7 , -7)

Find:

Perimeter of PQR

Computation:

Distance between two point = √(x1 - x2)² + (y1 - y2)²

Distance between PQ = √(-2 - 16)² + (-4 - 4)²

Distance between PQ = 18 unit

Distance between QR = √(16 - 7)² + (-4 + 7)²

Distance between QR = √81 + 9

Distance between QR = 9.48 unit (Approx.)

Distance between RP = √(7 + 2)² + (-7 + 4)²

Distance between RP = √81 + 9

Distance between RP = 9.48 unit (Approx.)

Perimeter of PQR = PQ + QR + RP

Perimeter of PQR = 18 + 9.48 + 9.48

Perimeter of PQR = 36.96

Perimeter of PQR = 37 units (Approx.)

8 0
3 years ago
A school is planning to construct two rectangular play areas in the playground. The length of play area A must be 1 foot longer
Anna11 [10]

Answer:

А.The system has two solutions, but only one is viable because the other results in a negative width.

Step-by-step explanation:

Given

Let:

L_A \to length of play area A

W_A \to width of play area A

L_B \to length of play area B

W_B \to width of play area B

x \to Area of A

y \to Area of B

From the question, we have the following:

L_A = 1 + 4W_A

W_B = 2 + W_A

L_B = 2 + 3W_B

x = y

The area of A is:

x = L_A * W_A

This gives:

x = (1 + 4W_A) * W_A

Open bracket

x = W_A + 4W_A^2

The area of B is:

y = L_B * W_B

y = (2 + 3W_B) * ( 2 + W_A)

Substitute: W_B = 2 + W_A

y = (2 + 3(2 + W_A)) * ( 2 + W_A)

Open brackets

y = (2 + 6 + 3W_A) * ( 2 + W_A)

y = (8 + 3W_A) * ( 2 + W_A)

Expand

y = 16 + 8W_A + 6W_A + 3W_A^2

y = 16 + 14W_A + 3W_A^2

We have that:

x = y

This gives:

W_A + 4W_A^2 = 16 + 14W_A + 3W_A^2

Collect like terms

4W_A^2 - 3W_A^2 + W_A  -14W_A  - 16 =0

W_A^2  -13W_A  - 16 =0

Using quadratic calculator, we have:

W_A = -14.1 or W = 1.13 --- approximated

But the width can not be negative; So:

W = 1.19

7 0
3 years ago
Solve for b:<br><br> b² = 8b + 84
levacccp [35]
B² = 8b + 84
b² - 8b - 84 = 0
b = <u>-(-8) +/- √((8)² - 4(1)(-84))</u> 
                       2(1)<u>
</u>b = <u>8 +/- √(64 + 336)</u>
                   2
b = <u>8 +/- √(400)
</u>               2<u>
</u>b = <u>8 +/- 20
</u>            2
b = 4 <u>+</u> 10
b = 4 + 10          b = 4 - 10
b = 14                b = -6
<u />
4 0
3 years ago
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