Ask question

Ask question

All categories

2 years ago

13

Quadrilateral EFGH is on a coordinate plane. Segment FG is on the line 3x − y = −2, and segment EH is on the 3x − y = −6. Which

statement proves how segments FG and EH are related?

1 answer:

liberstina [14]2 years ago

4 0

Answer:

the slope of both lines are the same.

Step-by-step explanation:

Given the following segment of the Quadrilateral EFGH on a coordinate Segment FG is on the line 3x − y = −2,

segment EH is on the 3x − y = −6.

To determine their relationship, we can find the slope of the lines

For line FG: 3x - y = -2

Rewrite in standard form y = mx+c

-y = -3x - 2

Multiply through by-1

y = 3x + 2

Compare

mx = 3x

m = 3

The slope of the line segment FG is 3

For line EH: 3x - y = -6

Rewrite in standard form y = mx+c

-y = -3x - 6

Multiply through by-1

y = 3x + 6

Compare

mx = 3x

m = 3

The slope of the line segment EH is 3

Hence the statement that proves their relationship is that the slope of both lines are the same.

You might be interested in

PLEASSS HELP!!!!!!! I REALLY NEED HELP!!

loris [4]

Answer:

$\underline{ \boxed{x = 11} } \\ \underline{ \boxed{ABE = 55}}$

Step-by-step explanation:

$if \: the \: question \: is \to \\ \overline{AZ} \: \perp \: \overline{CD}, \: m \angle{ABE} = 5x, \\ and \\ \: m \angle{EBD } = 3x + 2\\ but \: m \angle{EBD } + m \angle{ABE} = 90 \\ hence \to \\ (3x + 2) + (5x) = 90 \\ 8x + 2 = 90 \\ 8x = 90 - 2 \\ 8x = 88 \\ x = \frac{88}{8} \\ \boxed{ x = 11} \\ \\ if \: x = 11, \: then \to \\ m \angle{ABE} = \: 5x = 5(11) \\ \boxed{m\angle{ABE} =55}$

6 0

3 years ago

Find the center and the radius of the circle with the equation .

Simora [160]

We have been given an equation of circle $x^2+2x+y^2-2y-14=0$ . We are asked to find the center and radius of the circle.

First of all, we will write equation of circle is standard form $(x-h)^2+(y-k)^2=r^2$ , where center of circle is at point (h,k) and radius is r.

Let us complete the square for both x and y.

$x^2+2x+y^2-2y-14+14=0+14$

$x^2+2x+y^2-2y=14$

Adding half the square of coefficient of x and y term, we will get:

$(\frac{2}{2})^2=1^2=1$

$(\frac{-2}{2})^2=(-1)^2=1$

$x^2+2x+1+y^2-2y+1=14+1+1$

$(x+1)^2+(y-1)^2=16$

$(x-(-1))^2+(y-1)^2=4^2$

Therefore, the center of the circle is $(-1,1)$ and radius is 4 units.

4 0

3 years ago

F(x)=x^2-3x+15<br> find f(-8)

Vlada [557]

Answer:

103

Step-by-step explanation:

(-8)²-3(-8)+15

64+24+15

103

8 0

2 years ago

What is the measurement of angle POR

IgorLugansk [536]

Let us observe the given figure,

When two lines intersect each other, the angles opposite to each other are Vertically Opposite Angles. Vertically opposite angles are always equal in measure.

As, we can observe that the given lines intersect each other, and they form vertically opposite angles as $\angle POR , \angle SOQ$ and $\angle POS , \angle ROQ$

Therefore, $\angle POR = \angle SOQ$

Substituting the given measures of the angles, we get

$91-x^\circ = x+ 7^\circ$

$91-7 = x + x$

$84 = 2x$

$x = \frac{84}{2}$

So, x = $42^\circ$

Since, the measure of angle POR = $(91-x)^\circ$

= $(91-42)^\circ$

= $49^\circ$

Therefore, the measure of angle POR is 49 degrees.

8 0

4 years ago

I got 16/63 but i am not sure if its correct or not .

QveST [7]

Answer:

i think i got

sin x =16/65 not 16/63

3 0

3 years ago