The radius going through the tangent point is perpendicular to the tangent line, so ΔQRT is a right triangle.
QT²=QR²+RT²
QR and QS are both radii, so QR=QS=15
QT²=15²+36²
QT=39
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Step-by-step explanation:
Answer:
I'm gonna use the distance formula:
Distance = sqrt((x2-x1)^2 + (y2-y1)^2)
Distance = sqrt((6-3)^2 + (13-9)^2)
Distance = sqrt(3^2 + 4^2)
Distance = sqrt(9 + 16)
Distance = sqrt(25)
Distance = 5
The problem says every unit = 1 mile so 5 units = 5 miles.
Answer = 5 miles
Step-by-step explanation:
The system of equations has the solution:
x = 111/42
y = 27/14
z = -15/6
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How to solve the system of linear equations?</h3>
We start with 3 linear equations:
2x+y-z=8
x+4y+z=7
-3x+2y-3z=21
To solve this, first we need to isolate one of the variables. I will isolate x on the second one:
x = 7 - 4y - z
Now we can replace that in the other two:
2*( 7 - 4y - z) + y - z = 8
-3*(7 - 4y - z) + 2y - 3z = 21
Now we need to isolate other variable, let's isolate y on the above one:
14 - 8y - 2z + y - z = 8
-7y - 3z = 8 - 14 = -6
z = (-6 + 7y)/-3 = 2 - (7/3)*y
Now we replace this on the last equation:
-21 + 12y -3z + 2y - 3z = 21
14y - 6z = 42
14y - 6*(2 - (7/3)*y) = 42
14y - 12 + 14y = 42
28y = 42 + 12 = 54
y = 54/28 = 27/14
Now that we know the value of y, we can find the value of z:
z = 2 - (7/3)*y = 2 - (7/3)*27/14 = 2 - 27/6 = -15/6
And the value of x:
x = 7 - 4y - z = 7 - 4*(27/14) + 15/6 = 7 - 48/7 + 15/6 = 111/42
So the solution is:
x = 111/42
y = 27/14
z = -15/6
If you want to learn more about systems of equations:
brainly.com/question/13729904
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