1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
inn [45]
3 years ago
7

Gina wrote the following paragraph to prove that the segment joining the midpoints of two sides of a triangle is parallel to the

third side:
Given: ΔABC


Prove: The midsegment between sides Line segment AB and Line segment BC is parallel to side Line segment AC.


Draw ΔABC on the coordinate plane with point A at the origin (0, 0). Let point B have the ordered pair (x1, y1) and locate point C on the x-axis at (x2, 0). Point D is the midpoint of Line segment AB with coordinates at Ordered pair the quantity 0 plus x sub 1, divided by 2. The quantity 0 plus y sub 1, divided by 2 by the Distance between Two Points Postulate. Point E is the midpoint of Line segment BC with coordinates of Ordered pair the quantity x sub 1 plus x sub 2 divided by 2. The quantity 0 plus y sub 1 divided by 2 by the Distance between Two Points Postulate. The slope of Line segment DE is found to be 0 through the application of the slope formula: The difference of y sub 2 and y sub 1, divided by the difference of x sub 2 and x sub 1 is equal to the difference of the quantity 0 plus y sub 1, divided by 2, and the quantity 0 plus y sub 1, divided by 2, divided by the difference of the quantity x sub 1 plus x sub 2, divided by 2 and the quantity 0 plus x sub 1, divided by 2 is equal to 0 divided by the quantity x sub 2 divided by 2 is equal to 0 When the slope formula is applied to Line segment AC the difference between y sub 2 and y sub 1, divided by the difference of x sub 2 and x sub 1 is equal to the difference of 0 and 0, divided by the difference of x sub 2 and 0 is equal to 0 divided by x sub 2 is equal to 0, its slope is also 0. Since the slope of Line segment DE and Line segment AC are identical, Line segment DE and Line segment AC are parallel by the definition of parallel lines.


Which statement corrects the flaw in Gina's proof?


The slope of segments DE and AC is not 0.

Segments DE and AC are parallel by construction.

The coordinates of D and E were found using the Midpoint Formula.

The coordinates of D and E were found using the slope formula.
Mathematics
2 answers:
Viefleur [7K]3 years ago
8 0

Answer:

The coordinates of D and E were found using the Midpoint Formula.

Georgia [21]3 years ago
3 0

Answer: The answers are

(i) The slope of segments DE and AC is not 0.

(ii) The coordinates of D and E were found using the Midpoint Formula.

Step-by-step explanation:  We can easily see in the proof  that the co-ordinates of D and E were found using the mid-point formula, not distance between two points formula. So, this is the first flaw in the Gina's proof.

Also, we see that the slope of line DE and AC, both are same, not equal to 0 but is equal to

\dfrac{y_2}{x_2},

which is 0 only if y_2=0.

So, this is the second mistake.

Thus, the statements that corrects the flaw in Gina's proof are

(i) The slope of segments DE and AC is not 0.

(ii) The coordinates of D and E were found using the Midpoint Formula.

You might be interested in
Which calculations perform on the calculator will yelled in answer that is approximately the same as 76 ÷ 19/25? Check all that
Zinaida [17]

first:76 ÷ 19/25=100

a:80 / 20 * 1 / 25 =0.16

b:80 * 5/4 =100

c:80 / 5 ÷ 1 / 4 =64

d:80 * 5/4 =100

e:80 * 4/5 ​=64

So I think both B and D should be the right answer

6 0
3 years ago
A diameter of a circle has enpoints (-2, 10) and (6, -4) in the standard (x, y) coordinate plane. What is the center of the circ
wlad13 [49]

Answer:

The center of the circle is C(x,y) = (2, 3).

Step-by-step explanation:

The center of the circle is the midpoint of the segment between the endpoints. We can determine the location of the center by this vectorial expression:

C(x,y) = \frac{1}{2}\cdot R_{1}(x,y)+ \frac{1}{2}\cdot R_{2}(x,y) (1)

Where:

C(x,y) - Center.

R_{1} (x,y), R_{2} (x,y) - Location of the endpoints.

If we know that R_{1} (x,y) = (-2,10) and R_{2} (x,y) = (6,-4), then the location of the center of the circle is:

C(x,y) = \frac{1}{2}\cdot (-2,10)+\frac{1}{2}\cdot (6,-4)

C(x,y) = (-1, 5) + (3, -2)

C(x,y) = (2, 3)

The center of the circle is C(x,y) = (2, 3).

8 0
3 years ago
On a fishing excursion, jeannie keeps 1/4 of the fish she catches and releases the remainder of the fish back into the lake. if
postnew [5]
To find 1/3 of 3/4 we divide the fraction by 3 which is equal to 1/4 or alternatively we can multiply the denominator by 3 which gives us 3/12 (1/4=3/12)
So 3/12 of the realesed fish are trout

To find 2/3 of 1/4 we need to first divide 1/4 by 3 which is equal to multiplying the denominator times 3 which gives us 1/12 and then we multiply by 2 which gives us 2/12
So 2/12 of the kept fish are trout

To find the total amount we add 2/12 and 3/12
3/12+2/12= 5/12

ANSWER:
5/12 of the fish she catches are trout
7 0
3 years ago
Pleaaaaaaaaaaase hurry
ExtremeBDS [4]

Answer:

Lindsay owns 12 books

Step-by-step explanation:

1/3x:Lindsay

x:Destiny

1/4x: Henry

1/3x+1/4x+x=57

4/12x+3/12x+x=57

7/12x+x=57

19/12x=57

x=57*12/19

x=36

1/3x=36/3

1/3x=12

4 0
3 years ago
Read 2 more answers
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
3 years ago
Other questions:
  • Which equation demonstrates the distributive property?
    14·2 answers
  • How to convert 60miles into kilometers
    15·1 answer
  • -70+8p-4z identify the terms and the coefficient
    12·1 answer
  • Cardinal Corporation, a calendar year taxpayer, receives dividend income of $250,000 from a corporation in which it holds a 10%
    10·1 answer
  • A survey organization, as part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten y
    9·1 answer
  • Is -0.2222222... terminating or repeating
    6·2 answers
  • Please Help !!! :( I don't Understand !!!
    7·1 answer
  • My question is to write 32 in the form 2n
    7·2 answers
  • What is the domain of the piecewise function?
    5·1 answer
  • Converse of Pythagorean theorem, Find the missing side length of each triangle round to the nearest 10th if necessary.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!