Evaluate 3-5 using a number line.

Find the equation of the line passing through the point (–1, 5) and perpendicular to the line y = – 3x + 4.

dsp73

9514 1404 393

Answer:

A) 3y = x + 16

Step-by-step explanation:

The equation of a perpendicular line can be found by swapping the x- and y-coefficients and negating one of them. The constant will be chosen to match the given point.

Swapping coefficients, we get ...

-3y = x + c

Negating the y-coefficient gives ...

3y = x + c

Filling in the given point, we have ...

3(5) = -1 + c

16 = c

The equation of the perpendicular line can be written as ...

3y = x + 16 . . . . matches choice A

_____

Note that choice A is the only equation that gives a line with positive slope. The given equation has negative slope, so its perpendicular must have positive slope.

3 0

3 years ago

I need help ASAP 10 Points!!!

swat32

Purple color figure is similar to the original figure.

Solution:

Property of similar polygon:

Similar polygons have same shape, corresponding angles are congruent and the corresponding sides are in the same ratio.

(1) Compare original figure with green color figure:

Angles of the original figure:

37°, 53° and 90°.

Angles of the green color figure:

67°, 23° and 90°

These two angles of the triangles not congruent angles.

Hence these shapes are not similar.

(2) Compare original figure with yellow color figure:

Angles of the original figure:

37°, 53° and 90°.

Angles of the yellow color figure:

67°, 23° and 90°

These two angles of the triangles are not congruent angles.

Hence these shapes are not similar.

(3) Compare original figure with purple color figure:

Angles of the original figure:

37°, 53° and 90°

Angles of the purple color figure:

37°, 53° and 90°

These two triangles have same shape and congruent angles.

Ratio of the sides:

$$\frac{3}{6}=\frac{1}{2} , \ \frac{4}{8}=\frac{1}{2} , \ \frac{5}{10}=\frac{1}{2}$

These two triangles have same ratio.

Hence these shapes are similar.

Hence purple color figure is similar to the original figure.

4 0

3 years ago

HELP PLZ DUE TM!!!! 20 POINTS!!!

Flauer [41]

$\displaystyle\bf\\m \overset{\frown}{HE}=360-m \overset{\frown}{HL}-m \overset{\frown}{EV}-m \overset{\frown}{VL}\\m\overset{\frown}{HE}=360^o-40^o-130^o-110^o=360^o-280^o=80^o\\\\m\widehat{EYH}=m\widehat{EYV}=\frac{m \overset{\frown}{EV}-m\overset{\frown}{HE}}{2}=\frac{130^o-80}{2}=\frac{50^o}{2}=\boxed{\bf25^o}$