Y is less than or equal to 3/2x

A winery has a vat with two pipes leading to it. The inlet pipe can fill the vat in 5 hours, while the outlet pipe can empty it

Rashid [163]

Answer:

$time=40/3hours\approx 13.3hours$

Step-by-step explanation:

The rates are additive: you can calculate the inlet rate and the outlet rate and add them algebraically, i.e. the inlet rate will be positive and the outlet rate will be negative.

1. Inlet rate:

$1vat/5hours$

2. Outlet rate:

$1vat/8hours$

3. Net rate:

$\text{Inlet rate - outlet rate}=1vat/5hours-1vat/8hours\\\\ \text{Net rate}=(8-5)vat/40hour=3vat/40hour=(3/40)vat/hour$

4. Time to fill the vat

$rate=amount/time\implies time=amount/rate\\ \\ time=1vat/(3vat/40hour)$

$time=40/3hours\approx 13.3hours$

7 0

3 years ago

Answer as many as you can please (write in slope-intercept form)

RoseWind [281]

Answer: See below

Step-by-step explanation:

For the first one, we are already given our slope. All we need to do is find the y-intercept, b.

y=-2x+b

6=-2(-3)+b

6=6+b

b=0

The slope-intercept form is y=-2x.

For the second one, we need to first find the slope using $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$ .

$m=\frac{1-13}{3-(-6)} =\frac{-12}{9}$

Now that we have our slope, we can plug it into our slope-intercept form to solve for b.

$y=-\frac{12}{9} x+b$

$3=-\frac{12}{9}(1)+b$

$-\frac{9}{4} =b$

The slope-intercept form is $y=-\frac{12}{9} -\frac{9}{4}$ .

For the third one, we are already given the slope, so all we have to do is find b.

$y=-\frac{1}{2}x +b$

$-7=-\frac{1}{2} (-4)+b$

$-7=2+b$

$-9=b$

The slope-intercept form is $y=-\frac{1}{2} x-9$ .

For the last one, we need to first find the slope using $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$ .

$m=\frac{-8-2}{3-1}=\frac{-10}{2} =-5$

Now that we have our slope, we can plug it into our slope-intercept form and find b.

$y=-5x+b$

$2=-5(1)+b$

$2=-5+b$

$7=b$

Our slope-intercept form is $y=-5x+7$ .

4 0

3 years ago

How do I find the distance between (-2,5) and (22,12)

jok3333 [9.3K]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{22}~,~\stackrel{y_2}{12})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[22-(-2)]^2+[12-5]^2}\implies d=\sqrt{(22+2)^2+(12-5)^2} \\\\\\ d=\sqrt{24^2+7^2}\implies d=\sqrt{625}\implies d=25$

6 0

3 years ago

Read 2 more answers

10,000,000,000×200,000

Snowcat [4.5K]

10,000,000,000×200,000 = 2e+15

5 0

3 years ago

Read 2 more answers

The harbormaster wants to place buoys where the river bottom is 20 feet below the surface of the water. Complete the absolute va

Natalija [7]

Answer:

The answer is below

Step-by-step explanation:

The bottom of a river makes a V-shape that can be modeled with the absolute value function, d(h) = ⅕ ⎜h − 240⎟ − 48, where d is the depth of the river bottom (in feet) and h is the horizontal distance to the left-hand shore (in feet). A ship risks running aground if the bottom of its keel (its lowest point under the water) reaches down to the river bottom. Suppose you are the harbormaster and you want to place buoys where the river bottom is 20 feet below the surface. Complete the absolute value equation to find the horizontal distance from the left shore at which the buoys should be placed

Answer:

To solve the problem, the depth of the water would be equated to the position of the river bottom.

$d(h)=river \ bottom\\\\The \ river \ bottom=-20\ feet(below)\\\\d(h) = -20\\\\\frac{1}{5}|h-240|-48=-20\\ \\\frac{1}{5}|h-240|=-20+48\\\\\frac{1}{5}|h-240|=28\\\\|h-240|=28*5\\\\|h-240|=140\\\\h-240=140\ or\ h-240=-140\\\\h=140+240\ or\ h=-140+240\\\\h=380\ or\ h=100\\\\The\ buoys\ should\ be\ placed\ at\ 100\ feet\ and\ 380\ feet\ from\ left-hand\ shore$

4 0

3 years ago