Answer: 3
3x4 is 12
Hope this helps!
Answer:
see below
Step-by-step explanation:
The tangent function has been shifted upward by 2 units, but there has been no horizontal scaling. Any horizontal offset must be equal to some number of whole periods.
Choices A and B show tan( )+2, the correct vertical offset. However, choice A has a horizontal scale factor of 2. The correct choice is B, which has no horizontal scaling (the coefficient of x is 1) and a horizontal offset of π, one full period.
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<em>Comment on horizontal scaling</em>
Horizontal scaling is different from vertical scaling in that using k·x in place of x <em>compresses</em> the graph horizontally by a factor of k. On the other hand, using k·f(x) in place of f(x) <em>expands</em> the graph vertically by a factor of k.
Do u mean Linear? if so it is
The formula for the area of a triangle is
1/2 b•h
Let’s plug in our values !
1/2 (30•75)
1/2 • 2,250 = ?
2,250/2 = 1,125
The area of the triangle is 1,125 cm^2
I hope I helped !
Can you vote me brainliest, so I can level up? :)
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.
