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Natasha_Volkova [10]
4 years ago
14

If the distance from 0 to x on a number line is equal to 5, then –5 + x = 0

Mathematics
1 answer:
harkovskaia [24]4 years ago
6 0
In the problem -5+x=0,  x=5
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The sum of two positive numbers is 12. what two numbers will maximize the product g
son4ous [18]

Given two numbers x and y such that:

x + y = 12   ...    (1)


<span>two numbers will maximize the product g</span>

from  equation (1) 

y = 12 - x  

Using this value of y, we represent xy as

xy = f(x)= x(12 - x)

 f(x) = 12x - x^2

Differentiating the above function:

f'(x) = 12 - 2x

Maximum value of f(x) occurs at point for which f'(x) = 0.

Equating f'(x) to 0 we get:

12 - 2x = 0

 2x =  12

> x = 12/2 = 6

Substituting this value of x in equation (2):

y = 12 - 6 = 6

Therefore, value of xy is maximum when:

x = 6 and y = 6

The maximum value of xy = 6*6 = 36

4 0
3 years ago
Solve the following system of equations:
PSYCHO15rus [73]
If we multiply the first equation by 2 we get
2x - 4y = 12
but the second equation is
2x - 4y = 10

2x - 4y can't have 2 different values  so the are no solutions 
7 0
3 years ago
Read 2 more answers
A garden snail moves 1/6 foot in 1/3 hour. Find the speed of the snail in feet per hour
Arisa [49]
(1/6 ft)/ (1/3 hour)= 1/2 ft/hour

The final answer is 1/2 ft/hour~
6 0
3 years ago
I just wanted to see if I got my answers correct, please explain your answer so I can check up with mine.
kenny6666 [7]
B because you need to do 2/6 which is the probability of getting a number above 4 times 1/2 the probability of getting heads. 2/6 x 1/2 = 2/12 = 1/2
7 0
3 years ago
Read 2 more answers
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
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