Answer:
6 laps.
Step-by-step explanation:
On Monday she runs 2x + 8 miles, where x miles = the length of the trail.
On Tuesday she runs 4x + 4 miles and this is the same distance as Monday, so
2x + 8 = 4x + 4
4 = 2x
x = 2 miles which is the length of the trail.
So each day she runs 4(2) + 4 = 12 miles.
So on Saturday she runs 12/ 2 = 6 laps.
Answer:
2 cups of sugar for every 12 cookies
Step-by-step explanation:
131.7625 would be your answer here
The respective missing proofs are; Alternate interior; Transitive property; Converse alternate interior angles theore
<h3>How to complete two column proof?</h3>
We are given that;
∠T ≅ ∠V and ST || UV
From images seen online, the first missing proof is Alternate Interior angles because they are formed when a transversal intersects two coplanar lines.
The second missing proof is Transitive property because angles are congruent to the same angle.
The last missing proof is Converse alternate interior angles theorem
because two lines are intersected by a transversal forming congruent alternate interior angles, then the lines are parallel.
Read more about Two Column Proof at; brainly.com/question/1788884
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Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
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* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
