Answer:
(3, 6)
Step-by-step explanation:
If you create a mini graph, like I did, you would have seen that the middle point is at the co-ordinates (3,6). I have attached a image to this reply, with my "graph". (Sorry if its a little ugly, but that doesn't matter, the math does.)
Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.
<h3>
Inscribing a square</h3>
The steps involved in inscribing a square in a circle include;
- A diameter of the circle is drawn.
- A perpendicular bisector of the diameter is drawn using the method described as the perpendicular of the line sector. Also known as the diameter of the circle.
- The resulting four points on the circle are the vertices of the inscribed square.
Alicia deductions were;
Draws two diameters and connects the points where the diameters intersect the circle, in order, around the circle
Benjamin's deductions;
The diameters must be perpendicular to each other. Then connect the points, in order, around the circle
Caleb's deduction;
No need to draw the second diameter. A triangle when inscribed in a semicircle is a right triangle, forms semicircles, one in each semicircle. Together the two triangles will make a square.
It can be concluded from their different postulations that Benjamin is correct because the diameter must be perpendicular to each other and the points connected around the circle to form a square.
Thus, Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.
Learn more about an inscribed square here:
brainly.com/question/2458205
#SPJ1
Answer:
c = 1/7
Step-by-step explanation:
Here we have to solve the linear equation.
We need to find the value of c.
5 = 6 - 7c
Isolate the variable and constants
Add 7c on both sides, we get
7c + 5 = 6 - 7c + 7c
7c + 5 = 6
Subtract 5 from both sides, we get
7c + 5 - 5 = 6 - 5
7c = 1
Dividing both sides by 7, we get
c = 1/7
Thank you.
3(-2) -(1) -3(-2)
= -6 -1 +6 = -1
1a) False. A square is never a trapezoid. A trapezoid has only one pair of parallel sides while the other set of opposite sides are not parallel. Contrast this with a square which has 2 pairs of parallel opposite sides.
1b) False. A rhombus is only a rectangle when the figure is also a square. A square is essentially a rhombus and a rectangle at the same time. If you had a Venn Diagram, then the circle region "rectangle" and the circle region "rhombus" overlap to form the region for "square". If the statement said "sometimes" instead of "always", then the statement would be true.
1c) False. Any rhombus is a parallelogram. This can be proven by dividing up the rhombus into triangles, and then proving the triangles to be congruent (using SSS), then you use CPCTC to show that the alternate interior angles are congruent. Finally, this would lead to the pairs of opposite sides being parallel through the converse of the alternate interior angle theorem. Changing the "never" to "always" will make the original statement to be true. Keep in mind that not all parallelograms are a rhombus.
