HEEELP PLEASE

In the figure above, PQRS is a circle. If PQT and SRT<br>are straight lines, find the value of x.

Elena L [17]

Given:

PQRS is a circle, PQT and SRT are straight lines.

To find:

The value of x.

Solution:

Since PQRS is a circle, PQT and SRT are straight lines, therefore, PQRS isa cyclic quadrilateral.

We know that, sum of opposite angles of a cyclic quadrilateral is 180 degrees.

$m\angle SPQ+m\angle QRS=180^\circ$

$81^\circ+m\angle QRS=180^\circ$

$m\angle QRS=180^\circ-81^\circ$

$m\angle QRS=99^\circ$

Now, SRT is a straight line.

$m\angle QRT+m\angle QRS=180^\circ$ (Linear pair)

$m\angle QRT+99^\circ=180^\circ$

$m\angle QRT=180^\circ-99^\circ$

$m\angle QRT=81^\circ$ ...(i)

According to the Exterior angle theorem, in a triangle the measure of an exterior angle is equal the sum of the opposite interior angles.

Using exterior angle theorem in triangle QRT, we get

$m\angle PQR=m\angle QRT+m\angle QTR$

$x=81^\circ+22^\circ$

$x=103^\circ$

Therefore, the value of x is 103 degrees.

4 0

3 years ago

Simplify: (8^2)^3<br> please help:)

alexira [117]

Answer:

262144

Step-by-step explanation:

its exponents

6 0

3 years ago

Read 2 more answers

12/15 * 5/6

nydimaria [60]

La multiplicación de las fracciones es 2/3

Para multiplicar las dos fracciones tenemos que multiplicar en línea: numerador por numerador y denominador por denominador.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\sf \rightarrow \dfrac{12}{15} \times \dfrac{5}{6} }$

Resolvamos:

$\: \: \: {\sf \rightarrow\dfrac{12}{15}\times \dfrac{5}{6}=\dfrac{12 \times 5}{15 \times 6}=\dfrac{60}{90} =\dfrac{2}{3} }$

Por lo tanto, la multiplicación de las fracciones es 2/3

#Spj3

8 0

2 years ago

Read 2 more answers

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of

notsponge [240]

Answer:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, $(x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)$

<u>So, The coordinates of the triangle A'B'C' are:</u>

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

6 0

3 years ago

a student walks 50m on a bearing 0.25 degrees and then 200m due east how far is she from her starting point.

Inessa05 [86]

Answer:

Step-by-step explanation:

I'm going to use Physics here for this concept of vectors. Here are some stipulations I have set for the problem (aka rules I set and then followed throughout the problem):

** I am counting the 50 m as 2 significant digits even though it is only 1, and I am counting 200 as 3 significant digits even though it is only 1. 1 sig dig doesn't really give us enough accuracy, in my opinion.

** A bearing of .25 degrees is measured from the North and goes clockwise; that means that measured from the x axis, the angle is 89.75 degrees. This is the angle that is used in place of the bearing of .25 degrees.

** Due east has an angle measure of 0 degrees

Now let's begin.

We need to find the x and y components of both of these vectors. I am going to call the first vector A and the second B, while the resultant vector will be C. Starting with the x components of A and B:

$A_x=50cos(89.75)$ so