Brandon made $38.00 in 3 hours. How much would he make in 10 hours
2 answers:
Answer:
would make $126.67 in 10 hours
Step-by-step explanation:
To find how much Brandon will make in 10 hours, we will follow the steps below:
using proportion,
Let x be the amount he would make in 10 hours
$38 = 3 hours
x = 10 hours
cross-multiply
3×x =$38×10
3x = 380
Divide both-side of the equation by 3
=
At the left-hand side of the equation 3 will cancel-out 3 leaving us with just x while at the right-hand side of the equation 380 will be divided by 3, hence;
x ≈$ 126.67
He would make $126.67 in 10 hours
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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
===============================================
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
===============================================
The answer is: " 17.35 " .
__________________________________
→ The circumference of the circle is: " 17.35 units " .
__________________________________
The area, "A", of a circle:
A = * r² ;
Note: .
The area of the circumference, "C", of a circle:
C = 2 r .
__________________________________________
Given:
A = 24 ;
Use the formula for the area, "A" of a circle, to find the radius, "r" .
Then, plug the value obtained for "r" into the formula for the circumference, "C" , of a circle:
C = 2 r l
To solve for the circumference, "C", of the circle.
________________
A = ;
24 = 3.14 * r² ;
↔ 3.14 * r² = 24 ;
Divide each side of the equation by "(3.14)" ;
[3.14 * r²] / 3.14 = (24) / (3.14) ;
to get:
→ r² = 7.64331210191 ;
Take the positive square root of each side of the equation;
to isolate "r" on one side of the equation; & to solve for the radius, "r" ;
→ ⁺√(r²) = ⁺√(7.64331210191) ;
to get:
→ r = 2.76281034802 ;
____________________________
Now, plug this "obtained value" for the radius, "r" into the equation for the formula for the circumference, "C" of a circle:
→ C = 2 * r ;
to solve for the circumference, "C", of the circle:
→ C = 2 * (3.14) * (2.76281034802) ;
= (6.28) * (2.76281034802) ;
= 17.3504489856 ;
→ round to: " 17.35 units" .
__________________________________
The answer is: " 17.35 " .
__________________________________
→ The circumference of the circle is: " 17.35 units " .
__________________________________