Answer:
The probability that he teleports at least once a day = 
Step-by-step explanation:
Given -
Evan lives in Stormwind City and works as an engineer in the city of ironforge in the morning he has three Transportation options teleport ride a dragon or walk to work and in the evening he has the same three choices for his trip home.
Total no of outcomes = 3
P( He not choose teleport in the morning ) = 
P( He not choose teleport in the evening ) = 
P ( he choose teleports at least once a day ) = 1 - P ( he not choose teleports in a day )
= 1 - P( He not choose teleport in the morning )
P( He not choose teleport in the evening )
= 
= 
Please find diagram attached
Answer and explanation:
Perpendicular lines are lines that meet at 90 degrees/ right angle.
The question seems to be incomplete but I will explain what perpendicular lines are in the context of lines lll and mmm, with some assumptions about the lines.
Assume that line III is 30mm long and line mmm is 25mm long, we say that line III is perpendicular to line mmm if line III meets line mmm at 90 degrees as we see in the diagram attached.
This is a linear differential equation of first order. Solve this by integrating the coefficient of the y term and then raising e to the integrated coefficient to find the integrating factor, i.e. the integrating factor for this problem is e^(6x).
<span>Multiplying both sides of the equation by the integrating factor: </span>
<span>(y')e^(6x) + 6ye^(6x) = e^(12x) </span>
<span>The left side is the derivative of ye^(6x), hence </span>
<span>d/dx[ye^(6x)] = e^(12x) </span>
<span>Integrating </span>
<span>ye^(6x) = (1/12)e^(12x) + c where c is a constant </span>
<span>y = (1/12)e^(6x) + ce^(-6x) </span>
<span>Use the initial condition y(0)=-8 to find c: </span>
<span>-8 = (1/12) + c </span>
<span>c=-97/12 </span>
<span>Hence </span>
<span>y = (1/12)e^(6x) - (97/12)e^(-6x)</span>
Answer:
A) You would move it 2 times B) The Exponent would be 10^2
Step-by-step explanation:
This one's hard to explain..if you want you can message me about it :)
Answer:
Step-by-step explanation:
a). tan(75°) = 
= 
k = 
k = 2.947
k = 2.95 cm
b). cos(52°) = 
s = 
s = 25.988
s ≈ 25.99 cm
c). sin(5°) = 
= 
q = 
q = 184.727
q ≈ 184.73 cm