Several mountain climbers move up a mountain

Write an equation of the direct variation that includes the point (6,-2)

olga55 [171]

Answer:

$y = - \frac{1}{3}x$

Step-by-step explanation:

The equation of a direct variation is generally written as:

$y = mx$

Where m is the slope of the equation of the direct variation line.

We want a direct variation equation that contains (6,-2).

We substitute the x=6 and y=-2 to find m.

$- 2 = 6m$

Divide both sides by 6.

$\frac{ - 2}{ 6} = \frac{6m}{ 6}$

$- \frac{1}{3} = m$

The required equation is

$y = - \frac{1}{3}x$

7 0

3 years ago

Please help and good night

lana66690 [7]

Answer:

(7) n = ⁴/₃

(8) x = -5

(9) d = ³/₂

(10) v = -20

Step-by-step explanation:

(7) $n-\frac{8}{9} = \frac{4}{9}$

$n-\frac{8}{9} = \frac{4}{9} \\\\Collect \ similar \ terms \ together\\\\n = \frac{4}{9} + \frac{8}{9} \\\\n = \frac{4+8}{9} \\\\n = \frac{12}{9} \\\\n = \frac{4}{3}$

(8)

$\frac{3}{2} + \ x = -\frac{7}{2}\\\\collect \ similar \ terms \ together\\\\x = -\frac{7}{2} -\frac{3}{2} \\\\x = \frac{-7-3}{2} \\\\x = \frac{-10}{2} \\\\x = -5$

(9)

$\frac{6}{5} = \frac{d}{(\frac{5}{4} )} \\\\cross \ and \ multiply\\\\5d = 6(\frac{5}{4} )\\\\5d = \frac{30}{4} \\\\d = \frac{30}{4\times 5} \\\\d = \frac{30}{20} \\\\d = \frac{3}{2} \\\\d = 1\frac{1}{2}$

(10)

$-\frac{1}{8} v = \frac{5}{2} \\\\\frac{-v}{8} = \frac{5}{2} \\\\cross \ and \ multiply\\\\-2v = 40\\\\-v = \frac{40}{2} \\\\-v = 20\\\\v = -20$

6 0

3 years ago

Deceribe one kind of diagram you might draw to help you solve problem?

Gnoma [55]

You could draw out a word diagram

4 0

3 years ago

Graph the functions on the same coordinate axis. {f(x)=−2x+1g(x)=x2−2x−3

NikAS [45]

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. $f(x)=-2x+1$

If we compare this equation with the slope intercept form of a line which is given as

$y=mx+c$

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

$y=-2(1)+1$

$y=-2+1 = -1$

Let us find another point by putting x = 2 and solving it for y

$y=-2(2)+1$

$y=-4+1 = -3$

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

$y=-2(-2)+1$

$y=+1 = 5$

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) , (2,-3) and (-2,5) we joint them on line to obtain our line

2.

$g(x)=y=x^2-2x-3$

$y=x^2-2x+1-1-3$

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

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

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; $y=y=(0)^2-2(0)-3=0-0-3=-3$ ; (0,-3)

ii)x=-1 ; $y=(-1)^2-2(-1)-3=1+2-3=0$ ; (-1,0)

iii) x=2 ; $y=(2)^2-2(2)-3 = 4-4-3 =-3$ ;(2,-3)

iii) x=1 ; $y=(1)^2-2(1)-3 = 1-2-3 =-4$ ;(1,-4)

iii) x=-2 ; $y=(-2)^2-2(-2)-3 = 4+4-3 =5$ ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.

3 0

4 years ago

A sandbox has an area of 32 square feet, and

Ksivusya [100]

The width is 8 because 4x8=32

7 0

3 years ago