PLEASE HELP ME! SOMEONE!
1 answer:
The basic idea is that you draw perpendicular bisectors of two of the sides, then use their point of intersection as the center of the circle.
To construct the perpendicular bisectors, set your compass to a width that is greater than half the length of the longest side. In the attached picture, we noticed that DE is more than half the length of EF, so we simply used the length DE for the compass setting. Draw circles at each of the vertices with this radius.
There will be points where the circles intersect that are equidistant above and below the center of each side. Mark two pairs of these points. In the diagram, we have labelled them A, B, C, G. Draw the perpendicular bisectors of the triangle sides through these points. The points and lines are colored green in the figure.
The point of intersection of the bisectors is the center of the circle you want. In the figure, we have marked it H and colored it red. Reset your compass to have a radius of HD (or HE or HF—they should all be the same) and draw the circle at H through D, E, and F. It is red in the diagram. That is the circle the problem is asking for.
To construct the perpendicular bisectors, set your compass to a width that is greater than half the length of the longest side. In the attached picture, we noticed that DE is more than half the length of EF, so we simply used the length DE for the compass setting. Draw circles at each of the vertices with this radius.
There will be points where the circles intersect that are equidistant above and below the center of each side. Mark two pairs of these points. In the diagram, we have labelled them A, B, C, G. Draw the perpendicular bisectors of the triangle sides through these points. The points and lines are colored green in the figure.
The point of intersection of the bisectors is the center of the circle you want. In the figure, we have marked it H and colored it red. Reset your compass to have a radius of HD (or HE or HF—they should all be the same) and draw the circle at H through D, E, and F. It is red in the diagram. That is the circle the problem is asking for.
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Answer:
reciprocal = -14/15
= 15/-14
multiplicative inverse = ×
=
= 1
=
Answer:
<em>Maximum: (-1,9)</em>
Step-by-step explanation:
<u>Vertex form of the quadratic function</u>
If the graph of the quadratic function has a vertex at the point (h,k), then the function can be written as:
Where a is the leading coefficient.
We are given the following function:
To find the vertex, we need to complete squares. First, factor -2 on the first two terms:
The expression in parentheses must be completed to represent the square of a binomial. Adding 1 and subtracting 1:
Taking out the -1:
Factoring the trinomial and operating:
Comparing with the vertex form we have
Vertex (-1,9)
Leading coefficient: -2
Since the leading coefficient is negative, the function has a maximum value at its vertex, i.e.
Maximum: (-1,9)