All categories

2 years ago

David is performing the following construction. Based on the markings that are present, what construction is he performing?

Mathematics

2 answers:

lora16 [44]2 years ago

8 0

Answer:

He is bisecting the angle BAC.

Step-by-step explanation:

First he puts the compass point at point A and draws 2 small arcs at D and E. Then he places the compass point at D and then at E to form the 2 arcs that cross between the 2 line segments AB and AC.

The bisector is the line he can draw between A and the crossed arcs.

scZoUnD [109]2 years ago

3 0

Answer:

Bisecting an angle

Step-by-step explanation:

David first draw an angle CAB i.e $\angle CAB$

Then draw an arc cutting AC at D and AB at E .

With E as centre , draw an arc then with D as centre draw an arc

Both the arcs will intersect at some point .

If we join point A to this intersecting point , we get an angle bisector of $\angle CAB$

So, based on the markings that are present , he is actually bisecting the angle.

An angle bisector is a line or line segment which bisects an angle into two equal parts .

You might be interested in

Using the following information = Universal set = {2,3,4,5,6,7,8,9,12,13,14,16,20,22,56). Subset A = {9,12,13,20,22,56); Subset

Cloud [144]

$\dfrac{2}{5},\dfrac{13}{15},\dfrac{3}{5},\dfrac{1}{5},\dfrac{1}{5}$

step-by-step explanation:

The intersection of two sets A and B, denoted by A ∩ B, is the set containing all elements of A that also belong to B.

The union (denoted by ∪) of a collection of sets is the set of all elements in the collection.

$P(s)$ of a set $s$ is defined as ratio of number of elements in $s$ to the number of elements in $universal set$

given $Universalset=$ $\{$ $\text{2,3,4,5,6,7,8,9,12,13,14,16,20,22,56}$ $\}$

given $A=\{9,12,13,20,22,56\}$ and $H=\{4,5,8,9,16,22\}$

and $C=\{1,4,20,22,56\}$

For Question A:

$A$ ∩ $H$ = $\{9,12,13,20,22,56\}$ ∩ $\{4,5,8,9,16,22\}$

= $\{9,22}\}$

( $A$ ∩ $H$ )∪ $C$ = $\{9,22}\}$ ∪ $C=\{1,4,20,22,56\}$ = $\{1,4,9,20,22,56\}$

$p((A$ ∩ $H)$ ∪ $C)$ = $\frac{6}{15}=\frac{2}{5}$

For Question B:

$H$ ∩ $C$ = $\{4,22\}$

$p(H$ ∩ $C)$ '= $\frac{15-2}{15}=\frac{13}{15}$

For Question C:

$p(H)$ '= $\frac{15-6}{15}=\frac{3}{5}$

For Question D:

$C$ \ $H$ = $\{1,20,56\}$

$p(C$ \ $H)$ = $\frac{3}{15}=\frac{1}{5}$

For Question E:

$A$ \ $C$ = $\{9,12,13\}$

$p(A$ \ $C)$ = $\frac{3}{15}=\frac{1}{5}$

7 0

2 years ago

Please answer correctly !!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!

Olin [163]

U is a point in the line segment TV

TU + UV = TV

TU = 12

UV = 3

Hope this will help...

4 0

2 years ago

A rectangular window measures 54 inches by 24 inches. There is an a = 14-inch wiper blade attached by a b = 5-inch arm at the ce

nika2105 [10]

It the wiper blade (and arm) made a full rotation, 2 circles would be formed.

One small circle with radius b, and a larger circle with radius a+b.

The areas of these 2 circles are respectively:

$\pi b^2=\pi\cdot 5^2=25 \pi (in^2)$ and

$\pi (a+b)^2=\pi (14+5)^2=361\pi (in^2)$

120° is 1/3 of a complete angle 360° which form a circle, so the areas formed by the arm of length b, and the arm + the wiper blade (a+b) are:

$\frac{25 \pi}{3}$ and $\frac{361\pi}{3}$ in squared respectively.

the actual wiped area is $\frac{361\pi}{3}-\frac{25 \pi}{3}= \frac{336\pi}{3}=112\pi \approx351.7$ (in squared)

Area of the window is 54*24= 1296 (in squared)

thus, the ratio of the wiped area to the whole area of the window is 351.7/1296=0.271

Converted to percentages, this is 27.1%

Answer: 27.1%

3 0

2 years ago

What is the surface area of this rectangular prism?

morpeh [17]

All of the surfaces are rectangular

7 0

2 years ago

Read 2 more answers

HELP <br> multiple choice question

kompoz [17]

It's the third one because as Kendall spent more time training or running, their distance increased

5 0

2 years ago