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3 years ago

Given PQRS is a square, solve for x.

Mathematics

2 answers:

ivanzaharov [21]3 years ago

7 0

Answer:

x = 11

Step-by-step explanation:

The angles in a square are right , that is 90°

The diagonals bisect the angles , then

6x - 21 = 45 ( add 21 to both sides )

6x = 66 ( divide both sides by 6 )

x = 11

siniylev [52]3 years ago

6 0

Step-by-step explanation:

In the given figure, we can see that there is a square, and know that diagonals of squares bisect each other at 90° , so

Angle PTQ = 90°

( since they bisects each other at 90° )

now,

since the diagonals of the square are equal and they are bisected by each other so,

PT = TQ

then, Angle TPQ = Angle TQP

( since, PT = TQ , Opposite sides are equal )

In triangle PTQ ,

Angle TQP + Angle TPQ + Angle PTQ = 180°

( since, sum of all interior angles of a triangle is 180° )

=》Angle TQP + Angle TPQ + 90° = 180°

( PTQ = 90° )

=》2 × Angle TQP = 90°

( TPQ = TQP )

=》TQP = 45°

From here,

TQP = 45° = 6x - 21

=》6x - 21 = 45°

=》6x = 45 + 21

=》6x = 66

=》x = 66 / 6

=》x = 11

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Answer:

The vertex is the point (6,-31)

Step-by-step explanation:

we have

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therefore

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2 years ago

Can I get some help please? Thanks!

quester [9]

An ellipse (oval shape) is expressed by the following equation:

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}=1$ where h is the x coordinate of the center and k is the y coordinate of the center. Furthermore, a is the horizontal distance from the center, and b is the vertical distance from the center. Lastly, c is the distance from the center to one of the foci (they are spaced apart equally).

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Since the k value in this case is 0, the y value of both foci are 0. Also, since h and k are both 0, we know the center of the ellipse is at the origin.

So the foci are (-5, 0) and (5, 0)

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