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slega [8]
3 years ago
11

A quadrilateral has side lengths of 2x, (3x - 4), (2x - 3), and (x + 5). Find the perimeter.

Mathematics
2 answers:
Shalnov [3]3 years ago
7 0
Um i really do not know
babunello [35]3 years ago
4 0

Answer:

its fortnite

Step-by-step explanation:

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Can someone plz help me with this answer is not A I’m not understanding it plz help
STatiana [176]
I think it’s C. 136 ft squared
8 0
3 years ago
What is the solution of the system?
mario62 [17]
Y + 7 = 3x.....y = 3x - 7

6x - 2y = 12
6x - 2(3x - 7) = 12
6x - 6x + 14 = 12
14 = 12 (incorrect)

when ur variables cancel out and u r left with an incorrect statement, this means there is no solution because ur lines are parallel

so ur answer is : no solution
5 0
4 years ago
Read 2 more answers
Help asap will mark brainliest
katen-ka-za [31]

Answer:

(-5 3/4, 1 2/3)

Step-by-step explanation:

the line intersects right before the -6, so it’s not -6. yet, it is as close to -6 as you can get which would be -5 3/4. meaning the closes would be the third choice.

8 0
3 years ago
A computer programming team has 13 members. a. How many ways can a group of seven be chosen to work on a project? b. Suppose sev
Julli [10]

Answer:

1716 ;

700 ;

1715 ;

658 ;

1254 ;

792

Step-by-step explanation:

Given that :

Number of members (n) = 13

a. How many ways can a group of seven be chosen to work on a project?

13C7:

Recall :

nCr = n! ÷ (n-r)! r!

13C7 = 13! ÷ (13 - 7)!7!

= 13! ÷ 6! 7!

(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)

1235520 / 720

= 1716

b. Suppose seven team members are women and six are men.

Men = 6 ; women = 7

(i) How many groups of seven can be chosen that contain four women and three men?

(7C4) * (6C3)

Using calculator :

7C4 = 35

6C3 = 20

(35 * 20) = 700

(ii) How many groups of seven can be chosen that contain at least one man?

13C7 - 7C7

7C7 = only women

13C7 = 1716

7C7 = 1

1716 - 1 = 1715

(iii) How many groups of seven can be chosen that contain at most three women?

(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)

Using calculator :

(15 * 35) + (6 * 21) + (1 * 7)

525 + 126 + 7

= 658

c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?

(First in second out) + (second in first out) + (both out)

13 - 2 = 11

11C6 + 11C6 + 11C7

Using calculator :

462 + 462 + 330

= 1254

d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

Number of ways with both in the group = 11C5

Number of ways with both out of the group = 11C7

11C5 + 11C7

462 + 330

= 792

8 0
3 years ago
7(4x - 2) - 4(2x - 8)
vesna_86 [32]

Answer:

20x + 18

Step-by-step explanation:

We need to use the distributive property, where we essentially take the sum of the product of the outside number with each of the inside terms.

In 7(4x - 2), 7 is the outside number and 4x and -2 are the inside numbers, so:

7(4x - 2) = 7 * 4x + 7 * (-2) = 28x - 14

In 4(2x - 8), 4 is the outside number and 2x and -8 are the inside numbers, so:

4(2x - 8) = 4 * 2x + 4 * (-8) = 8x - 32

Now, we have:

28x - 14 - (8x - 32) = 28x - 14 - 8x + 32 = 20x + 18

The answer is 20x + 18.

6 0
3 years ago
Read 2 more answers
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