20 POINTS!! ASAP!!! PLEASE!! Use the figure to answer questions 1 and 2. In each diagram, line k is parallel to line l, and line
t intersects lines k and I.
1 answer:
Answer:
see the explanation
Step-by-step explanation:
we know that
<u><em>Alternate Exterior Angles</em></u> are created where a transversal crosses two lines. Notice that the two alternate exterior angles are equal in measure if the two lines are parallel
In this problem
----> by alternate exterior angles
One way to verify alternate exterior angles is to see that they are the vertical angles of the alternate interior angles
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The steps involving properties of equality are shown in painful detail.
For Step 4, you add 7 to both sides of the equation (with the goal of cancelling the -7 on the left). So you have
6x -7 +7 = 12 +7 . . . . . . . showing 7 added to both sides
In the next step, this is simplified to the way you might have written it to start with:
6x = 19
There should be no confusion in any of this. All solutions of all equations make use of the properties of equality like this. Addition is used to cancel subtraction, division is used to cancel multiplication. (Root-taking cancels powers, and vice-versa; inverse trig functions cancel trig functions; and so on.)
For Step 4, you add 7 to both sides of the equation (with the goal of cancelling the -7 on the left). So you have
6x -7 +7 = 12 +7 . . . . . . . showing 7 added to both sides
In the next step, this is simplified to the way you might have written it to start with:
6x = 19
There should be no confusion in any of this. All solutions of all equations make use of the properties of equality like this. Addition is used to cancel subtraction, division is used to cancel multiplication. (Root-taking cancels powers, and vice-versa; inverse trig functions cancel trig functions; and so on.)
Jacob wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn,so he needs no fence on that side.
Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn).
one side of the length is not counted for perimeter because one side of length will be against the barn.
Perimeter = 400 ft
Perimeter of rectangle = L + W + W
400 = L + 2W
L = 400 - 2W
Area = L * W
Replace L by 400 - 2W
A(W) = (400 - 2W) * W
Now we find out x coordinate of vertex to find the width that maximize the area
a= -2 and b = 400
The width w would maximize the area is w = 100ft
To find maximum area we plug in 100 for W in A(W)
the maximum area is 20,000 square feet