Which statement is a example of a theorem? a line is an infinite set of points that extends in opposite directions,

In the figure above, quadrilateral ABCD is a parallelogram. Let x represent the measure of angle GBF, y represent the measure of

ira [324]

Answers:
measure angle x = 40°
measure angle y = 35°
measure angle z = 55°

Explanation:
Part (a): getting angle x:
In triangle BED, we have:
measure angle BED = 90°
measure angle BDE = 50°
Therefore:
measure angle DBE = 180 - (90+50) = 40°
Now, we have angle DBE and angle GBF vertically opposite angles.
This means that they are both equal. Therefore angle GBF = 40°
Since angle GBF is x, therefore:
x = 40°

Part (b): getting angle y:
We know that the sum of measures of angles on a straight line is 180.
This means that:
angle GBF + angle GBC + angle CBE = 180
We have:
angle GBF = 40°
angle GBC = 105°
angle CBE = y
Therefore:
40 + 105 + y = 180
y = 35°

Part (c): getting angle z:
In triangle BCE, we have:
measure angle BCE = z
measure angle BEC = 90°
measure angle CBE = 35°
Therefore:
z + 90 + 35 = 180
z = 55°

Hope this helps :)

8 0

3 years ago

Line g passes through points (5, 9) and (3, 2). Line h passes through points (9, 10) and (2, 12). Are line g and line h parallel

icang [17]

For this case we find the slopes of each of the lines:

The g line passes through the following points:

$(x_ {1}, y_ {1}) :( 3,2)\\(x_ {2}, y_ {2}) :( 5,9)$

So, the slope is:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {9-2} {5-3} = \frac {7} {2}$

Line h passes through the following points:

$(x_ {1}, y_ {1}) :( 9,10)\\(x_ {2}, y_ {2}) :( 2,12)$

So, the slope is:

$m = \frac {y_ {2} -y_ {1}}{x_ {2} -x_ {1}} = \frac {12-10} {2-9} = \frac {2} {- 7} = - \frac {2} {7}$

By definition, if two lines are parallel then their slopes are equal. If the lines are perpendicular then the product of their slopes is -1.

It is observed that lines g and h are not parallel. We verify if they are perpendicular:

$\frac {7} {2} * - \frac {2} {7} = \frac {-14} {14} = - 1$

Thus, the lines are perpendicular.

Answer:

The lines are perpendicular.

8 0

3 years ago

Use the graph of f below. Assume the entire function is graphed below. Find the domain

enot [183]

ANSWER

The domain of the function is

$- 3 \leqslant x \leqslant 6$

or

[-3,6]

EXPLANATION

The domain of a function refers to the values of x for which the function is defined.

From the graph, we can see that the function begins at (-3,6) and ends at (6,-4).

Therefore the function is defined for x=-3 to x=6.

Hence the domain of the function is

$- 3 \leqslant x \leqslant 6$

or

[-3,6]

5 0

3 years ago

someone, please help me asap, that would be great, offering 15 points, and please do not attach a link for the answer, thanks.

algol13

Pretty sure the answer is 288

7 0

3 years ago

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |subtract $y$ from both sides

$\dfrac{2}{3}x-y+b=0$ |multiply both sides by 3

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |multiply both sides by $\sqrt{13}$

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |divide both sides by 3

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

4 0

3 years ago