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lara [203]
3 years ago
10

Find the perimeter of a quadrilateral with vertices at C (−1, 2), D (−2, −1), E (2, −2), and F (1, 1). Round your answer to the

nearest hundredth when necessary.
Mathematics
1 answer:
expeople1 [14]3 years ago
7 0
To get the perimeter of the quadrilateral, the distance formula can be used. We simply have to subtract the y coordinate of one point to another and the same for the x coordinate. The square root of the sum of the squares of the difference is the distance.The results can then be added to get the perimeter. So,
CF = sqrt ( (2-(-1))^2 + (-1-(-2))^2 ) = sqrt(10)
FE = sqrt ( (-2-1)^2 + (2-1)^2 ) = sqrt(10)
ED = sqrt ( (-1-(-2))^2 + (-2-2)^2 ) = sqrt(17)
DC = sqrt ( (2-(-1)^2 + (-1-(-2))^2 ) = sqrt(10)

P = CF + FE + ED + DC = sqrt(10) + sqrt(10) + sqrt(17) + sqrt(10) = 13.61

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What is 513 + _ + 130 = 764? what is the missing number and can u pls do a explantation
dmitriy555 [2]

Answer:

121

Step-by-step explanation:

The missing number is 121. I got this by doing 764-130-513=121. You would do this because you are trying to find the number that would allow 130 and 513 to add up to 764.

Another way to look at this problem is 513+130= 643

764-643=121

In conclusion, 513+121+130=764

4 0
3 years ago
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He sum of two consecutive integers is â71â71. find the two integers
Crank
The two consecutive integers can be represented by x and x+1.

x+(x+1)=71

Combine like terms
2x+1=71

Subtract 1 from both sides
2x=70

Divide both sides by 2
x=35

Final answer: 35, 36
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3 years ago
Find the mid point or these coordinates
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(2,1) is the correct answer
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Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks. Two students in t
melomori [17]

Answer:

The system of equation used to find the Total number of flash cards made are \left \{ {{5x+5} \atop {6x}} \right..

In <u>5 minutes</u>, each student will have <u>30</u> flash cards.

Step-by-step explanation:

Given:

Number of flash cards already made by Tristan = 5

Number of flash card Tristan makes in 1 min = 5

Number of flash card Josh makes in 1 min = 6

We need to find the How long will Josh have same number of flashcards.

Also We need to find how many flashcards will each student have at that point.

Solution:

Let the number of minutes be'x'.

Now we can say that;

Total number of flash cards made by Tristan is equal to Number of flash cards already made by Tristan plus Number of flash card Tristan makes in 1 min multiplied by number of minutes.

framing in equation form we get;

Total number of flash cards made by Tristan = 5+5x

Also We can say that;

Total number of flash cards made by Josh is equal Number of flash card Josh makes in 1 min multiplied by number of minutes.

framing in equation form we get;

Total number of flash cards made by Josh = 6x

Hence the system of equation used to find the Total number of flash cards made are \left \{ {{5x+5} \atop {6x}} \right..

Now to find the number of minutes when both would have same number of flash cards we will make both the equation equal.

so we get;

5+5x=6x

On solving we get;

Combining like terms we get;

6x-5x=5\\\\x=5\ mins

Hence It would take 5 minutes when both will have same number of flash cards.

To find the number of flash cards each would had at that time we will substitute value of x=5 in the system of equations.

so we get;

number of flash cards made by Tristan = 5x+5 = 5\times5+5=25+5=30

number of flash cards made by Josh = 6x =6\times 5 =30

Hence In <u>5 minutes</u>, each student will have <u>30</u> flash cards.

6 0
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