Plzzz answer Parts A B and C for BRILLIANT answer

Write a linear equation. Explain it in number please! 20 points!!!

juin [17]

To find a linear question, we must have two points on the line and we must know the slope.

Let us choose two points at random:

(-6,12) and (-5,14)

Now let us find the slope of the linear line:

$m = \frac{y2-y1}{x2-x1} =\frac{14-12}{-5--6} = 2$

So the slope is 2.

Now let us set up the equation which now requires only one point and the slope

let us use the point (-6,12) and the slope of 2

$y-12 = 2(x+6)\\y = 2x +12 +12\\y= 2x +4$

So the linear line is y = 2x + 4.

Hope that helps!

8 0

2 years ago

What is the measure of minor arc BD?

Vaselesa [24]

Answer:

$minor\ arc\ BD=100^o$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

A <u><em>circumscribed angle</em></u> is the angle made by two intersecting tangent lines to a circle

so

In this problem

BC and CD are tangents to the circle

BC=CD ----> by the Two Tangent Theorem

That means

Triangle ABC and Triangle ADC are congruent

so

$m\angle BAC=m\angle DAC$

Find the measure of angle BAC

In the right triangle ABC

$m\angle BAC+m\angle BCA=90^o$

substitute given value

$m\angle BAC+40^o=90^o$

$m\angle BAC=90^o-40^o=50^o$

Find the measure of angle BAD

$m\angle BAD=2m\angle BAC$

$m\angle BAD=2(50^o)=100^o$

Find the measure of minor arc BD

we know that

$minor\ arc\ BD=m\angle BAD$ -----> by central angle

therefore

$minor\ arc\ BD=100^o$

8 0

3 years ago

Read 2 more answers

How many triangles and quadrilaterals altogether can be formed using the vertices of a 7-sided regular polygon?

MAVERICK [17]

Answer: The correct option is

(E) 70.

Step-by-step explanation: We are given to find the number of triangles and quadrilaterals altogether that can be formed using the vertices of a 7-sided regular polygon.

To form a triangle, we need any 3 vertices of the 7-sided regular polygon. So, the number of triangles that can be formed is

$n_t=^7C_3=\dfrac{7!}{3!(7-3)!}=\dfrac{7\times6\times5\times4!}{3\times2\times1\times4!}=35.$

Also, to form a quadrilateral, we need any 4 vertices of the 7-sided regular polygon. So, the number of quadrilateral that can be formed is

$n_q=^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!\times3\times2\times1}=35.$

Therefore, the total number of triangles and quadrilaterals is

$n=n_t+n_q=35+35=70.$

Thus, option (E) is CORRECT.

8 0

3 years ago

16 Which expression is represented by the model shown below?

aleksklad [387]

I think it’s the first one

5 0

2 years ago

An arithmetic sequence with a third term of 8 and a constant difference of 5

Vanyuwa [196]

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ \hrulefill\\[0.5em] a_3=8\\ n=3\\ d=5 \end{cases} \\\\\\ a_3=a_1+(3-1)5\implies 8=a_1+(2)5 \\\\\\ 8=a_1+10\implies -2=a_1 \\\\\\ \begin{cases} a_1=-2\\ d=5 \end{cases}\implies a_n=-2+(n-1)d$

5 0

3 years ago