Based on the table of values below find the slope between points where x=2 and where x=6

write the standard form of the line that is parallel to 2 x + 3 y = 4 and passes through the point (1, -4).

jolli1 [7]

Answer:

2x +3y = -10

Step-by-step explanation:

The given line equation is in standard form, so the one we write will match that form. A parallel line will have the same x- and y-coefficients, but a different constant on the right.

Put the point values in the left-side expression to find the required right-side constant: (x, y) = (1, -4)

2(1) +3(-4) = constant = 2-12 = -10

Your line is ...

2x +3y = -10

7 0

3 years ago

This graph shows the solution to which inequality?

wariber [46]

Answer:

B. y > 2/3x + 1

Step-by-step explanation:

To find slope we'll use the following formula,

$\frac{y^2-y^1}{x^2-x^1}$

(-3,-1) (3,3)

3 - -1 = 4

3 - -3 = 6

2/3x

The y intercept is 1,

we know this because that's the point the line touches the y axis.

Thus,

the answer is B. y > 1/3x + 1.

Hope this helps :)

6 0

4 years ago

The parking garage charges $3 for the first hour of parking and $1.45 for each additional hour.

Alika [10]

I don't know what you are asking but my best guess is it will be
$4.46 in total

8 0

3 years ago

Read 2 more answers

Find the volume of a cone with a radius of 3 millimeters and a 7 height of millimeters. Use 22/7 for 3.14 .

wolverine [178]

Answer:

66

Step-by-step explanation:

So the volume of a cone is $\pi r^2*\frac{h}{3}$

r = 3

h = 7

$9\pi *\frac{7}{3} = 21\pi$

$\frac{22}{7} *21=66$

6 0

3 years ago

A function $f$ has a horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (a

Masja [62]

1. $f$ has a horizontal asymptote at $y=-4$

This means that

$\displaystyle\lim_{x\to\pm\infty}f(x)-(-4)=0$

(for at least one of these limits)

2. $f$ has a vertical asymptote at $x=3$

This means that $f$ has a non-removable discontinuity at $x=3$ . Since $f$ is some rational function, there must be a factor of $x-3$ in its denominator.

3. $f$ has an $x$ -intercept at (1, 0)

This means $f(1)=0$ .

(a) With

$f(x)=\dfrac{ax+b}{x+c}$

the second point above suggests $c=-3$ . The first point tells us that

$\displaystyle\lim_{x\to\pm\infty}\frac{ax+b}{x-3}+4=0=\lim_{x\to\pm\infty}\frac{ax+b+4x-3}{x-3}=0$

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if $a=-4$ so that the linear terms vanish.