Answer:
70
Step-by-step explanation:
An angle formed by 2 tangent lines is equal to one half the measure of the major arc minus the measure of the minor arc.
we are given the minor arc ( 110 ) however we are not given the major arc.
Because circles make a 360 degree rotation we can find the major arc by subtracting the measure of the minor arc from 360
Thus, major arc = 360 - 110 = 250
Now that we have identified the major and minor arc we can find x
Recall that x = 1/2 ( major arc - minor arc )
Thus, x = 1/2 ( 250 - 110 )
250 - 110 = 140
140 / 2 = 70
Hence, x = 70
Answer:
-1
Step-by-step explanation:
in numeracy numbers on right hand are bigger than numbers on left hand -1 is bigger than all negative integers. they all fall on the left of -1
Answer:
8kg
Step-by-step explanation:
4m=2kg
16m=?
=16m×2
=32
=32÷4
=8kg
Hope this will help ya!
Answer:
- hits the ground at x = -0.732, and x = 2.732
- only the positive solution is reasonable
Step-by-step explanation:
The acorn will hit the ground where the value of x is such that y=0. We can find these values of x by solving the quadratic using any of several means.
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<h3>graphing</h3>
The attachment shows a graphing calculator solution to the equation
-3x^2 + 6x + 6 = 0
The values of x are -0.732 and 2.732. The negative value is the point where the acorn would have originated from if its parabolic path were extrapolated backward in time. Only the positive horizontal distance is a reasonable solution.
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<h3>completing the square</h3>
We can also solve the equation algebraically. One of the simplest methods is "completing the square."
-3x^2 +6x +6 = 0
x^2 -2x = 2 . . . . . . . . divide by -3 and add 2
x^2 -2x +1 = 2 +1 . . . . add 1 to complete the square
(x -1)^2 = 3 . . . . . . . . written as a square
x -1 = ±√3 . . . . . . . take the square root
x = 1 ±√3 . . . . . . . add 1; where the acorn hits the ground
The numerical values of these solutions are approximately ...
x ≈ {-0.732, 2.732}
The solutions to the equation say the acorn hits the ground at a distance of -0.732 behind Jacob, and at a distance of 2.732 in front of Jacob. The "behind" distance represents and extrapolation of the acorn's path backward in time before Jacob threw it. Only the positive solution is reasonable.