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dezoksy [38]
3 years ago
6

What percent of 86 is 50 to the nearest whole percent

Mathematics
1 answer:
seropon [69]3 years ago
5 0

Answer:

58%

Step-by-step explanation:

50/86 is approximately equal to 0.58, and when you convert that into percentage you get 58%

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A number that is being divided.
olga55 [171]

Answer:

5

Step-by-step explanation:

12-7=5

3 0
3 years ago
assuming that tide duration from low tide to high tide is 6 hours and 25 minutes, going from a high of 3.21ft and down to a low
mixer [17]
To answer this you will need to write the ratio of difference in the tides over the amount of time that goes by.

1. 3.21 ft - 0.20 ft= 3.01 ft change.
2. 3.01 change over 6 5/12 hr.
3. Divide these to find the lapse rate per hour. This is about 0.47 ft per hour.
4 0
3 years ago
What is this answer -v+5+6v=1+5v+3
vivado [14]

Answer:

no solution

Step-by-step explanation:

Given

- v +5 + 6v = 1 + 5v + 3 ← combine like terms on both sides

5v + 5 = 5v + 4 ( subtract 5 from both sides )

5v = 5v - 1 ( subtract 5v from both sides )

0 = - 1 ← not possible

Hence the equation has no solution

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=-3x%5E%7B2%7D%20-4y%5E%7B2%7D%20-z%5E%7B2%7D%2B6xy-6x%2B4z" id="TexFormula1" title="-3x^{2} -4
vovikov84 [41]
What are the directions??
8 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
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