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miss Akunina [59]
3 years ago
11

The points (–2, 4) and (–2, –2) are vertices of a heptagon. Explain how to find the length of the segment formed by these endpoi

nts. How long is the segment?
Mathematics
2 answers:
spin [16.1K]3 years ago
8 0

Answer:

Sample Response: The segment is vertical because the x-coordinates are the same. To find the distance, you can count the spaces between the points. The segment is 6 units long.

Step-by-step explanation:

Scrat [10]3 years ago
5 0

Answer:

The segment is 6 units long.

Step-by-step explanation:

The points (–2, 4) and (–2, –2) are vertices of a heptagon. We have to explain how to find the length of the segment formed by these endpoints.

If two points at the ends of a straight line PQ are P(x_{1},y_{1}) and Q(x_{2},y_{2}), then the length of the segment PQ will be given by the formula  

\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}

Now, in our case the two points are (-2,4) and (-2,-2) and the length of the segment will be  

\sqrt{(- 2 - (- 2))^{2} + (4 - (- 2))^{2}} = 6 units. (Answer)

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Answer:

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Step-by-step explanation:

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plug in the values:

1.4/0.2 + 2(0.2) - (1.4)^2 = 7 + 0.4 - 1.96 = 5.44

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again, plug in the values:

2[-0.75-2(-1)]^3 - 0/12.71 = 2[1.25]^3 - 0 = 2[1.95] = 3.9

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Read 2 more answers
Solve the system by using a matrix equation (Picture provided)
Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

10x - 6y =2

it can be represented in matrix form as\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]

A= \left[\begin{array}{cc}7&-4\\10&-6\end{array}\right]

X= \left[\begin{array}{c}x\\y\end{array}\right]

B= \left[\begin{array}{c}4\\2\end{array}\right]

i.e, AX=B

or X= A⁻¹ B

A⁻¹ = 1/|A| * Adj A

determinant of A = |A|= (7*-6) - (-4*10)

                                    = (-42)-(-40)

                                    = (-42) + 40 = -2

so, |A| = -2

Adj A=  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]

A⁻¹ =  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]/ -2

A⁻¹ =  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]

X= A⁻¹ B

X=  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]

X= \left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]

X= \left[\begin{array}{c}12-4\\20-7\end{array}\right]

X= \left[\begin{array}{c}8\\13\end{array}\right]

x= 8, y= 13

solution set= (8,13).

Option b is correct.

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3 years ago
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