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san4es73 [151]
3 years ago
11

Which theorems or postulates could be used to show ABC =ECD

Mathematics
1 answer:
Kamila [148]3 years ago
3 0

Answer:

?

Step-by-step explanation:

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Need help with this question.
daser333 [38]

Answer:

1. Positive

2. Negative

3. Negative

4. Positive

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
5x + 2y = 6 3x + y = 4 Which of the following is part of the solution to the system of equations?
BabaBlast [244]
5x + 2y = 6
3x + y = 4....multiply by -2
-------------
5x + 2y = 6
-6x - 2y = -8 (result of multiplying by -2)
-------------add
-x = -2
x = 2

5x + 2y = 6
5(2) + 2y = 6
10 + 2y = 6
2y = 6 - 10
2y = -4
y = -4/2
y = -2 <=== here it is
6 0
4 years ago
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Solve: -5y- 9 = -(y - 1)
RoseWind [281]

Answer:

Exact Form:

y=-\frac{5}{2}

6 0
3 years ago
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2. Draw what might be the missing figure in
adoni [48]

Answer:

<em><u>PLEASE </u></em><em><u>ATTACH</u></em><em><u> </u></em><em><u>THE </u></em><em><u>FIGURE </u></em><em><u>LIKE </u></em><em><u>THIS.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

8 0
2 years ago
For what values of h are the vectors [Start 3 By 1 Matrix 1st Row 1st Column negative 1 2nd Row 1st Column 4 3rd Row 1st Column
Bogdan [553]

Answer:

The set is linearly dependent for every value of h.

Step-by-step explanation:

Recall that a set of vector {a,b,c,d} is a linearly dependent set of vectors if any of the vectors can be written as a linear combination of the other ones.

We are given the following vectors.a=(-1,4,6), b=(5,2,-3), c=(3,-5,-4), d=(12,-20, h). At first, we will check if a,b,c are linearly independent. To do so, we will calculate the following determinant (the procedure of the calculation is omitted).

\left|\begin{matrix}-1 & 5 & 3 \\ 4 & 2 & -5 & \\ 6 & -3 & -4\end{matrix}\right| = -119\neq 0

Since the determinant is not zero, this implies that the vectors a,b,c are all linearly independent. Since a,b,c are all vectors in \mathbb{R}^3 which is a 3-dimensional space, and they are 3 linear independent vectors, then they are automatically a base of this space. Consider now the vector d. Since {a,b,c} is a base of \mathbb{R}^3, then it generates any vector of this space(i.e any other vector of the space is a linear combination of {a,b,c}). In particular, d. So the set {a,b,c,d} is linearly independent for any value of h

3 0
3 years ago
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