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2 years ago

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: b² = 8b + 84

levacccp [35]

B² = 8b + 84
b² - 8b - 84 = 0
b = -(-8) +/- √((8)² - 4(1)(-84))
 2(1)
b = 8 +/- √(64 + 336)
 2
b = 8 +/- √(400)
 2
b = 8 +/- 20
 2
b = 4 + 10
b = 4 + 10 b = 4 - 10
b = 14 b = -6

4 0

3 years ago