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2 years ago

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Mathematics

1 answer:

ella [17]2 years ago

5 0

Answer:

A non-equilateral rhombus.

Step-by-step explanation:

We can solve this graphically.

We start with square:

ABCD

with:

A = (11, - 7)

B = (9, - 4)

C = (11, - 1)

D = (13, - 4)

Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:

AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13

BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13

CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13

DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13

And we change C by C' = (11, 1)

In the image you can see the 5 points and the figure that they make:

The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.

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x^2+10x-2=0

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Answer:

1) f⁻¹(x) = 6 ± 2√(x -1)

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Step-by-step explanation:

In general, swap x and y, then solve for y. Quadratics, as in the first problem, do not have an inverse function: the inverse relation is double-valued, unless the domain is restricted. Here, we're just going to consider these to be "solve for ..." problems, without too much concern for domain or range.

1) x = f(y)

x = (1/4)(y -6)² +1

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Answer:

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Step-by-step explanation:

Given the equation system:

$\left \{ {{3x+y+4z=-3} \atop {-x+y+4z=17}} \right.$

We obtain the following matrix:

$\left[\begin{array}{cccc}3&1&4&-3\\-1&1&4&17\end{array}\right]$

Step 1: Multiply the fisrt row by 1/3.

$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\-1&1&4&17\end{array}\right]$

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$\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&\frac{4}{3} &\frac{16}{3}&16\end{array}\right]$

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$\left \{ {{x=-5} \atop {y=12-4z}} \right.$

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