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3 years ago

A line passes through the points (9, 30) and (18, 30). Which statement is true about the line?

Mathematics

2 answers:

icang [17]3 years ago

6 0

Answer:

It has a slope of zero because the change in the y-values is 0.

Step-by-step explanation:

asambeis [7]3 years ago

3 0

Answer:

the line is horizontal

Step-by-step explanation:

Note the y- coordinates of the 2 points the line passes through are equal.

This indicates that the line is horizontal and parallel to the x- axis

The equation of a horizontal line is y = c

where c is the value of the y- coordinates the line passes through.

Here the y- coordinates are 30, hence

Equation of horizontal line is y = 30

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Dmitrij [34]

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3 years ago

The slope of the line containing the points (6, 4) and (-5, 3) is: A. 1/11 B. 1 C. -1

svetoff [14.1K]

$\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-5}~,~\stackrel{y_2}{3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{3-4}{-5-6}\implies \cfrac{-1}{-11}\implies \cfrac{1}{11}$

5 0

3 years ago

Read 2 more answers

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

Alternative method

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

3 years ago

Find the volume of the cone, using 3.14 for π Radius= 10 Height= 6

Zigmanuir [339]

Answer:

V≈628.32

Step-by-step explanation:

7 0

3 years ago

The equation of a straight line is y = 20 – X.

Reptile [31]

Given:

The equation of a straight line is

$y=20-x$

The x value of a point on the line is 14.

To find:

The y value of this point.

Solution:

If a point (x,y) lines on line, then the equation of line must be stratified by that point.

We have, the equation of a straight line.

$y=20-x$

Putting x=14, we get

$y=20-14$

$y=6$

Therefore, the y-value is 6 when the x-value is 14.

3 0

3 years ago