What needs to be corrected in the following construction for copying ∠ABC with point D as the vertex?
Answers
To construct an angle bisector for ∠ABC, use a compass and straightedge to create intersecting arcs, connecting the vertex with the intersection point to form the bisector. Avoid changing compass width or inaccurate arc intersections.
Constructing an angle bisector for ∠ABC using only a compass and a straightedge involves a series of precise steps. Here are the instructions along with common mistakes to avoid:
Step-by-Step Instructions:
Draw ∠ABC: Begin by drawing the angle ∠ABC with the given vertex at point B.
Place Compass at Point B: Use your compass and place its needle point (the sharp end) at point B, the vertex of the angle.
Adjust Compass Width: Open the compass to a width that allows you to draw two arcs that intersect both rays of the angle. Ensure the compass width remains fixed during the construction.
Draw Arcs: With the compass set, draw an arc that intersects the first ray, AB. Keep the compass needle fixed at B, and draw another arc that intersects the second ray, BC. Label the points of intersection with the rays as D and E.
Connect B and E: Using your straightedge, draw a straight line that connects point B and point E.
Bisect the Angle: The line BE bisects angle ∠ABC, creating two equal angles, ∠ABE and ∠EBC. Angle ∠ABE is the bisector of ∠ABC.
Common Mistakes to Avoid:
Changing Compass Width: Keeping the compass width consistent is crucial. Changing it during the construction will result in an inaccurate angle bisector.
Inaccurate Arc Intersection: Ensure that the arcs drawn from points B intersect the rays AB and BC accurately at points D and E. Inaccurate intersection points will lead to an incorrect angle bisector.
Not Labeling the Bisector: It's essential to label the constructed angle bisector (∠ABE) to distinguish it from the original angles.
By following these steps carefully and avoiding common mistakes, you can accurately construct an angle bisector for ∠ABC using only a compass and a straightedge.
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The question probable may be:
How do you construct an angle bisector for ∠ABC using only a compass and a straightedge? Provide step-by-step instructions, and highlight any common mistakes to avoid in the construction.
From looking at both the pictures shown for the problems which can be found in the attachment below. I concluded that what needs to be corrected in the following construction for copying ∠ABC with point D as the vertex is answer choice:
C) The third arc should cross the second arc.
I hope this helps, Regards.
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A function f (x)=-8x^2. What is f (-3)?
B.2.06
C.2.5
D.3.2
E.12.8
Answers
Answer: D. 3.2
Step-by-step explanation:
Given : In parallelogram LMNO,
LM = 4.12, MN = 4, LN = 5, and OM = 6.4.
Diagonals and intersect at point R.
We know that diagonals of a parallelogram bisect each other.
Since R is the intersection point of both diagonals.
⇒R is the mid point of OM.
Thus OR=
Therefore, OR=3.2
Answers
Let's start with the easy one: the square
The perimeter of a square = 4x where x = one side of the square
Since the width of the shape is 5, that means the other three sides of the square should also be five so:
perimeter of square = 4(5)
perimeter of square = 20 units
Now let's find the perimeter of the semicircle.
That equation is this:
perimeter of semicircle = 1/2(π x diameter) + diameter
We know that the width is 5 which means the diameter is 5.
Let's plug that in!
perimeter of semicircle = 1/2(π x 5) + 5
Multiplying pi by 5 and dividing it in half will give us a decimal, so we have to round that to the second place:
perimeter of semicircle = 1/2(5π) +5
perimeter of semicircle ≈ 7.85 + 5
perimeter of semicircle ≈ 12.85 units
Like we said at the beginning, this is two shapes in one so we have to add the perimeter of the square to the perimeter of the semicircle
perimeter of square = 20 units
perimeter of semicircle ≈ 12.85 units
20 + 12.85 = 32.85 units
So the perimeter of the shape = 32.85 units
B. (30+6)/6+3*2^2
C.40-[5*(3+1)+8]
D.[(3+5)*2-5]*3
Answers
A. 15+(8*3-6)
inside the parentheses, we multiply first
15+(24-6)
now simplify the rest of the parentheses
15+18
33
so answer "A'' equals 33
B. (30+6)/6+3*2^2
we simplify the inside of the parentheses
36/6+3*2^2
we simplify the exponent
36/6+3*4
6+12
18
So NO, answer "B" is not correct
C. 40-[5*(3+1)+8]
all the way on the inside of the parentheses,
40-[5*4+8]
multiplying comes before adding, so
40-[20+8]
inside the brackets,
40-28
12
So answer "C" is also incorrect
D. [(3+5)*2-5]*3
inside the parentheses
[8*2-5]*3
multiplying before subtracting
[16-5]*3
11*3
33
So the only answers that equal 33 are answer "A" and answer "D" equal 33
Answers
Answer:
Using the Pythagorean theorem for isosceles right triangle, the length of the hypotenuse is sqrt( 2 x 10^2) = ~14.1 in
Hope this helps!
:)
Answers
Answer:
would that be 12
Step-by-step explanation:
Answers
Answer:
57
Step-by-step explanation:
5y - 33 = 57.
Add 33.
5y = 90
Divide by five.
y = 18
5(18) - 33 = 57