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pickupchik [31]
3 years ago
13

EXPERTS HELP!! Jane was playing a game with her sister, Sally. Jane was down by 7. After her next turn, she was down by 3. Which

of the following mathematical statements represents this scenario?
A. - 7 + -3 = -10
B.
- 7 + 3 = - 4
C. - 7+ 4 = - 3
D. - 7 + - 4 = - 11
Mathematics
2 answers:
cricket20 [7]3 years ago
7 0

Answer:

A. - 7 + -3 = -10 this is your answer

max2010maxim [7]3 years ago
6 0
............Te he answer is C
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Trigonometric relationship

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Subract the childs age in years from 30 then divide the result by 2
Alona [7]

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8pm?  try that

Step-by-step explanation:

3 0
3 years ago
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Systems of equations with different slopes and different y-intercepts have one solution?​
Artemon [7]

Step-by-step explanation:

The systems of linear equations can have:

1. No solution: When the lines have the same slope but different y-intercept. This means that the lines are parallel and never intersect, therefore, the system of equations has no solution.

2. One solution: When the lines have different slopes and intersect at one point in the plane. The point of intersection will be the solution of the system

3. Infinitely many solutions: When the lines have the same slope and the y-intercepts are equal. This means that the equations represents the same line and there are infinite number of solution.

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6 0
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6 points are place on the line a, 4 points are placed on the line b. How many triangles is it possible to form such that their v
yawa3891 [41]

Answer: 96

Step-by-step explanation:

Ok, lines a and b are parallel.

We can separate this problem in two cases:

Case 1: 2 vertex in line a, and one vertex in line b.

Here we use the relation:

"In a group of N elements, the total combinations of sets of K elements is given by"

C = \frac{N!}{(N - K)!*K!}

Here, the total number of points in the line is N, and K is the ones that we select to make the vertices of the triangle.

Then if we have two vertices in line a, we have:

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C = \frac{6!}{4!*2!}  = \frac{6*5}{2} = 3*5 = 15

And the other vertex can be on any of the four points on the line b, so the total number of triangles is:

C = 15*4 = 60.

But we still have the case 2, where we have 2 vertices on line b, and one on line a.

First, the combination for the two vertices in line b is:

We use N = 4 and K = 2.

C = \frac{4!}{2!*2!} = \frac{4*3}{2} = 6

And the other vertice of the triangle can be on any of the 6 points in line a, so the total number of triangles that we can make in this case is:

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Then, putting together the two cases, we have a total of:

60 + 36 = 96 different triangles

3 0
3 years ago
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