2 years ago

Solutions to the inequality below? 10 ≤ 2 x − 1 ≤ x + 5

Mathematics

1 answer:

Georgia [21]2 years ago

6 0

Answer:

$10 \leqslant 2x - 1 \leqslant x + 5 \\ 10 + 1 \leqslant 2x - x \leqslant 5 + 1 \\ 11 \leqslant x \leqslant 6$

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4 years ago

Prove the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus. use the distance formula

riadik2000 [5.3K]

In order to prove this, we have to put the trapezoid to the coordinate system. In the attached photo you can see how it has to be put. The coordinates for the vertices of trapezoid written according to the midpoint principle. By using the distance between two points formula, we can find the coordinates for the vertices of the rhombus.
$D_{x} = \frac{-2b-2a}{2}=-b-a$ and $D_{y}= \frac{2c+0}{2}=c$ . The coordinates of D is $(-b-a, c)$
$E_{x} = \frac{-2b+2b}{2}=0$ and $E_{y}= \frac{2c+2c}{2}=2c$ . The coordinates of E is $(0,2c)$
Since we have the reflection in this graph, the coordinates of F is $(b+a, c)$
And the coordinates of G is (0,0).

Using the distance formula, we can find that
$DE= \sqrt{(-b-a)^{2}+ c^{2}}=\sqrt{(b+a)^{2} + c^{2}}$
$EF= \sqrt{(b+a)^{2} + c^{2} }$
$GF= \sqrt{(b+a )^{2} + c^{2} }$
$DG= \sqrt{(b+a)^{2}+ c^{2}$

Since all the sides are equal this completes our proof. Additionally, we can find the distances of EG and DF in order to show that the diagonals of this rhombus are not equal. So that it is not a square, but rhombus.