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2 years ago

The figure shows a construction completed by hand.

1 answer:

7 0

Based on the construction, we can logically deduce that: A. yes; the compass was kept at the same width to create the arcs for points C and D.

<h3>What is a line segment?</h3>

A line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points and it typically has a fixed length.

In Geometry, a line segment can be measured by using the following measuring instruments:

A scale (ruler)

A divider

A compass

In Geometry, an arc can be defined as a trajectory that is generally formed when the distance from a given point has a fixed numerical value.

Based on the construction with arcs created above and below the line segment from points A, we can infer and logically deduce that it is true that the compass was kept at the same width to create the arcs for points C and D.

In conclusion, yes, the construction demonstrated how to bisect a line segment correctly by hand.

Read more on arcs here: brainly.com/question/11126174

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Complete Question:

The construction has a given segment AB. Arcs have been created above and below the segment from points A that are equidistant from point A. The compass was kept at the same distance, placed on point B, and two additional arcs were created above and below the segment that intersect with the first arcs created. The intersection of the arcs above the segment created point C. The intersection of the arcs below the segment created point D. A line was drawn from point C to D through the segment.

Does the construction demonstrate how to bisect a segment correctly by hand?

A. Yes; the compass was kept at the same width to create the arcs for points C and D.

B. Yes; a straightedge was used to create segment CD.

C. No; the compass was not kept at the same width to create the arcs for points C and D.

D. No; a straightedge was used to create segment CD.

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Two triangles have bases with lengths 5in and 8in.

Elis [28]

Answer:

i. The ratio of the areas of the two triangles is 5:8.

ii. The area of the larger triangle is 24 in².

Step-by-step explanation:

Let the area of the smaller triangle be represented by $A_{1}$ , and that of the larger triangle by $A_{2}$ .

Area of a triangle = $\frac{1}{2}$ x b x h

Where; b is its base and h the height.

Thus,

a. The ratio of the area of the two triangles is:

$\frac{area of the smaller triangle}{area of the larger triangle}$

Area of smaller triangle = $\frac{1}{2}$ x b x h

= $\frac{1}{2}$ x 5 x h

= $\frac{5}{2}$ h

Area of the lager triangle = $\frac{1}{2}$ x b x h

= $\frac{1}{2}$ x 8 x h

= 4h

So that;

Ratio = $\frac{\frac{5}{2}h }{4h}$

= $\frac{5}{8}$

The ratio of the areas of the two triangles is 5:8.

b. If the area of the smaller triangle is 15 in², then the area of the larger triangle can be determined as;

$\frac{area of the smaller triangle}{area of the larger triangle}$ = $\frac{5}{8}$

$\frac{15}{A_{2} }$ = $\frac{5}{8}$

5 $A_{2}$ = 15 x 8

= 120

$A_{2}$ = $\frac{120}{5}$

= 24

The area of the larger triangle is 24 in².

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3 years ago

Which of the following is not an example of a recursive sequence?

wolverine [178]

A recursive pattern is a sequence of numbers that entails a pattern between succeeding numbers. in this case, we can see that sequence 2 has an arithmetic difference of 9, that is the proceeding terms are 40, 49, 58.. The third one has also an arithmetic difference that is increasing by 3 each. The answer then is A which has no pattern.

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3 years ago

-1/2(2x-6)+9=-8 answer

Oxana [17]

Answer:

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3 years ago

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What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

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3 years ago