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Leni [432]
2 years ago
13

Please someone help me .I need help. Focus on number 4

Mathematics
1 answer:
eimsori [14]2 years ago
7 0

Answer:

No

Step-by-step explanation:

To determine if the point (- 3, 5) is a solution to both equations.

Substitute the x and y values into the left side of both equations and if equal to the right sides then the point is a solution.

x - 3y = - 3 - 3(5) = - 3 - 15 = - 18 ← True

- 2x + y = - 2(- 3) + 5 =  6 + 5 = 11 ← False

Thus (- 3, 5) is not a solution to the system of equations.

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JulsSmile [24]
This is the factor tree check oit the pic to find out

6 0
2 years ago
Help please this is my last one then I’m done and o get to do whatever I want because I’m done with my homework but I need this
Alexxandr [17]

Answer:

The answer is 4 + 3/6 + 2/6

Step-by-step explanation:

3 0
3 years ago
vanessa drew a figure and said that it had an infinite number of lines of symmetry. what figure did she draw?
Korolek [52]
A circle
That's a very easy one cuz if u try u will find it to be correct

5 0
3 years ago
Read 2 more answers
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
2 years ago
The length of a rectangular garden is 3yd more than twice it’s width. The perimeter of the garden is 36yd. What are the width an
Basile [38]
2L + 2W = 36
2(3+2W) + 2W = 36
6 + 4W + 2W = 36
6W + 6 = 36
        -6  = -6
6W=30
W= 30/6 = 5     L= 3 + 2W = 3 +2 (5) = 3 +10= 13

Width is 5
Length is 13


7 0
3 years ago
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