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3 years ago

Is the line y = 3x – 7 parallel or perpendicular to 3x + 9y = 9? Explain your answer.

2 answers:

7 0

Y = 3x - 7...slope here is 3 and y int is -7

3x + 9y = 9
9y = -3x + 9
y = -1/3x + 1....slope here is -1/3 and y int is 1

the slopes 3 and -1/3 are negative reciprocals of each other...therefore, ur lines are perpendicular

vampirchik [111]3 years ago

4 0

Answer:

The lines are perpendicular because perpendicular lines have slopes that are opposite reciprocals. To compare the slopes, you can rewrite the second equation in slope-intercept form as y = (–1/3)x + 1. The slope of the given line is 3, and the slope of the second line is –1/3. These have a product of –1, which means they are opposite reciprocals. Thus, the two lines are perpendicular.

Step-by-step explanation:

sample response :)

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3 0

4 years ago

Please explain how you got the answer also, thank you

harina [27]

Answer:

See Explanation

Step-by-step explanation:

You are doing simple math within the square root. For instance where x=-2 you substitute -2 for x then add 2 which gives you 0 leaving you with -6.

for x=2 you end up with "square root 2+2" which gives you 4 and the square root of 4 is 2 so you add -6 to 2 giving you the answer of -4.

7 0

3 years ago

I have an assignment and I am having trouble with it. Can someone please help ASAP???

bezimeni [28]

Answer:

A) Find the sketch in attachment.

In the sketch, we have plotted:

- The length of the arena on the x-axis (90 feet)

- The width of the arena on the y-axis (95 feet)

- The position of the robot at t = 2 sec (10,30) and its position at t = 8 sec (40,75)

The origin (0,0) is the southweast corner of the arena. The system of inequalities to descibe the region of the arena is:

$0\leq x \leq 90\\0\leq y \leq 95$

Since the speed of the robot is constant, it covers equal distances (both in the x- and y- axis) in the same time.

Let's look at the x-axis: the robot has covered 10 ft in 2 s and 40 ft in 8 s. There is a direct proportionality between the two variables, x and t:

$\frac{10}{2}=\frac{40}{8}$

So, this means that at t = 0, the value of x is zero as well.

Also, we notice that the value of y increases by $\frac{75-30}{8-2}=7.5 ft/s$ (7.5 feet every second), so the initial value of y at t = 0 is:

$y(t=0)=30-7.5\cdot 2 =15 ft$

So, the initial position of the robot was (0,15) (15 feet above the southwest corner)

The speed of the robot is given by

$v=\frac{d}{t}$

where d is the distance covered in the time interval t.

The distance covered is the one between the two points (10,30) and (40,75), so it is

$d=\sqrt{(40-10)^2+(75-30)^2}=54 ft$

While the time elapsed is

$t=8 sec-2 sec = 6 s$

Therefore the speed is

$v=\frac{54}{6}=9 ft/s$

The equation for the line of the robot is:

$y=mx+q$

where m is the slope and q is the y-intercept.

The slope of the line is given by:

$m=\frac{75-30}{40-10}=1.5$

Which means that we can write an equation for the line as

$y=mx+q\\y=1.5x+q$

where q is the y-intercept. Substituting the point (10,30), we find the value of q:

$q=y-1.5x=30-1.5\cdot 10=15$

So, the equation of the line is

$y=1.5x+15$

By prolonging the line above (40,75), we see that the line will hit the north wall. The point at which this happens is the intersection between the lines

$y=1.5x+15$

and the north wall, which has equation

$y=95$

By equating the two lines, we find:

$1.5x+15=95\\1.5x=80\\x=\frac{80}{15}=53.3 ft$

So the coordinates of impact are (53.3, 95).

The distance covered between the time of impact and the initial moment is the distance between the two points, so:

$d=\sqrt{(53.5-0)^2+(95-15)^2}=95.7 ft$

From part B), we said that the y-coordinate of the robot increases by 15 feet/second.

We also know that the y-position at t = 0 is 15 feet.

This means that the y-position at time t is given by equation:

$y(t)=15+7.5t$

The time of impact is the time t for which

y = 95 ft

Substituting into the equation and solving for t, we find:

$95=15+7.5t\\7.5t=80\\t=10.7 s$

The path followed by the robot is sketched in the second graph.

As the robot hits the north wall (at the point (53.3,95), as calculated previously), then it continues perpendicular to the wall, this means along a direction parallel to the y-axis until it hits the south wall.

As we can see from the sketch, the x-coordinate has not changed (53,3), while the y-coordinate is now zero: so, the robot hits the south wall at the point

(53.3, 0)

The perimeter of the triangle is given by the sum of the length of the three sides.

- The length of 1st side was calculated in part F: $d_1 = 95.7 ft$

- The length of the 2nd side is equal to the width of the arena: $d_2=95 ft$

- The length of the 3rd side is the distance between the points (0,15) and (53.3,0):

$d_3=\sqrt{(0-53.3)^2+(15-0)^2}=55.4 ft$

So the perimeter is

$d=d_1+d_2+d_3=95.7+95+55.4=246.1 ft$

The area of the triangle is given by:

$A=\frac{1}{2}bh$

where:

$b=53.5 ft$ is the base (the distance between the origin (0,0) and the point (53.3,0)

$h=95 ft$ is the height (the length of the 2nd side)

Therefore, the area is:

$A=\frac{1}{2}(53.5)(95)=2541.3 ft^2$

The percentage of balls lying within the area of the triangle traced by the robot is proportional to the fraction of the area of the triangle with respect to the total area of the arena, so it is given by:

$p=\frac{A}{A'}\cdot 100$