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Yuliya22 [10]
2 years ago
13

Consider the graphs of j(x)== sin (6x- pi 2 )+11 and k(x) = 7sin(6x - pi/2) - 11 Which feature of the graph of k(x) is different

from the graph of j(x) ? the amplitude the frequency the phase shift the vertical shift
Mathematics
1 answer:
Tcecarenko [31]2 years ago
4 0

Answer:

The answer is D

You typed the first equation wrong but i just did this on edgenuity (with the right equations) and got it right

Step-by-step explanation:

You might be interested in
Find all polar coordinates of point P = (6, 31°).
wel

as far as I can tell, is just a matter of going around the circle many or infinite times around.

so 6,31° is the first point, the next point will be one-go-around, 6, 31+360 => 6, 391°

then the next will be 6, 391+360 => 6, 751° and so on.

so we can say is (6, 31° ±360°n), n ∈ ℤ.

6 0
2 years ago
If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen
Lunna [17]
Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation

Calculate z-scores for the following random variable and determine their probabilities from standard tables.

x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088

x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912

x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587

x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413

Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36

Answer: 27

Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36

Answer: 27

Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78

Answer: 204

Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150

Answer: 150

3 0
2 years ago
Make w the subject of the formula y-aw=2w-1
kipiarov [429]
Answer: <span>w = [ y + 1] / [a + 2]

Solution step by step:
</span>
1) given <span>formula: y-aw=2w-1

2) transpose aw and - 1

2w + aw = y + 1

3) common factor w:

w (a + 2) = y + 1

4) divide both sides by (a + 2):

w = [ y + 1] / [a + 2]
</span>
6 0
2 years ago
Consider the following system of equations: 10 + y = 5x + x2 5x + y = 1 The first equation is an equation of a . The second equa
aleksley [76]

Answer: The first equation is an equation of a parabola. The second equation is an equation of a line.

Explanation:

The first equation is,

10+y=5x+x^2

In this equation the degree of y is 1 and the degree of x is 2. The degree of both variables are not same. Since the coefficients of y and higher degree of x is positive, therefore it is a graph of an upward parabola.

The second equation is,

5x+y=1

In this equation the degree of x is 1 and the degree of y is 1. The degree of both variables are same. Since both variables have same degree which is 1, therefore it is linear equation and it forms a line.

Therefore, the first equation is an equation of a parabola. The second equation is an equation of a line.

5 0
2 years ago
Read 2 more answers
A rectangular piece of land is 40m long and 25m wide. A path of uniform width and 426 m^2 area sorrounds
fredd [130]

Answer:

Width of the uniform path that surrounds the piece of land = 3 m

Step-by-step explanation:

Let the width of the path that surrounds the piece of land be x.

The situation described is sketched in the attached image to this solution.

The dimension of the Length of the piece of land including the uniform path that surrounds it = (40 + 2x) m

The Breadth of the piece of land including the uniform path that surrounds it = (25 + 2x) m

The area of the piece of land including the uniform path that surrounds it

= (Area of the piece of land) + (Area of the uniform path that surrounds it)

Area of the piece of land = Length × Breadth = 40 × 25 = 1000 m²

Area of the uniform path that surrounds it = 426 m² (Given in the question)

The area of the piece of land including the uniform path that surrounds it = 1000 + 426 = 1426 m²

But the area of the piece of land including the uniform path that surrounds it is also

= (Length of the piece of land including the uniform path that surrounds it) × (Breadth of the piece of land including the uniform path that surrounds it

= (40 + 2x) + (25 + 2x)

= 1000 + 80x + 50x + 4x²

= (4x² + 130x + 1000) m²

Equation these 2 areas

4x² + 130x + 1000 = 1426

4x² + 130x - 426 = 0

Solving the quadratic equation

x = 3 or -35.5

Since the width cannot be negative,

x = 3 m

Hope this Helps!!!

7 0
2 years ago
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