Solve the equation for the problem above

Look at the Schoology problem provided below.

lys-0071 [83]

Answer:

k = 4 units

Step-by-step explanation:

Using Pythagoras' identity on the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

k² + 7.5² = 8.5²

k² + 56.25 = 72.25 ( subtract 56.25 from both sides )

k² = 16 ( take the square root of both sides )

k = $\sqrt{16}$ = 4

6 0

3 years ago

Read 2 more answers

What is X? x divided by 4/9 = 9/10

Damm [24]

Answer:

2/5

Step-by-step explanation:

$\frac{x}{\frac{4}{9} } =\frac{9}{10}$

$(\frac{9}{10})(\frac{4}{9} )=\frac{36}{90}$

36 divided by 90 is 2/5.

When solving for X you do the opposite to a number if you are moving it to the other side of the problem. so, since x is being divided by 4/9, 4/9 would be multiplied to the opposite or to the 9/10.

7 0

3 years ago

1. A sine function has the following key features:

andrew11 [14]

Problem 1

See the attached image (figure 1)

16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3

The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3

===============================================

Problem 2

See the attached image (figure 2)

T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0

The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1

===============================================

Problem 3

See the attached image (figure 3)

Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3

===============================================

Problem 4

See the attached image (figure 4)

a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline

The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)

===============================================

Problem 5

See the attached image (figure 5)

a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline

The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20

===============================================

8 0

3 years ago

Read 2 more answers

The manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one Monday. The

zavuch27 [327]

The standard deviation of the frequency distribution is 5.54

<h3>How to determine standard deviation?</h3>

The table of values is given as:

x f(x)

0-3 13

4-7 13

8-11 10

12-15 11

16-19 0

20-23 3

Rewrite the table by calculating the class midpoints:

x f(x)

1.5 13

5.5 13

9.5 10

13.5 11

17.5 0

21.5 3

Start by calculating the mean using:

$\bar x = \frac{\sum fx}{\sum f}$

This gives

$\bar x = \frac{1.5 * 13 + 5.5 * 13 + 9.5 * 10 + 13.5 * 11 + 17.5 * 0 + 21.5 * 3}{13 + 13 + 10 + 11 + 0 +3}$

Evaluate

$\bar x = 7.98$

The standard deviation is then calculated as:

$\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f }}$

So, we have:

$\sigma= \sqrt{\frac{13 * (1.5 - 7.98)^2 + 13 * (5.5 - 7.98)^2 + (9.5 - 7.98)^2 * 10 + (13.5 - 7.98)^2 * 11 + (17.5 - 7.98)^2 * 0 + (21.5 - 7.98)^2 * 3}{13 + 13 + 10 + 11 + 0 +3}}$