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3 years ago

Convert the flow rate of 250 gallons per minute (gpm) to a flow rate of gallons per second.

Mathematics

2 answers:

Blizzard [7]3 years ago

7 0

Answer:

multiply by

Convert from Convert to

US gpd US gpm cfm IMP gpd IMP gpm

m3/s 22800000 15852 2119 19000000 13200

m3/min 380000 264.2 35.32 316667 220

m3/h 6333.3 4.403 0.589 5277.8 3.67

liter/sec 22800 15.852 2.119 19000 13.20

liter/min 380 0.2642 0.0353 316.7 0.22

liter/h 6.33 0.0044 0.00059 5.28 0.0037

US gpd 1 0.000695 0.000093 0.833 0.000579

US gpm 1438.3 1 0.1337 1198.6 0.833

cfm 10760.3 7.48 1 8966.9 6.23

Imp gpd 1.2 0.00083 0.00011 1

Step-by-step explanation:

googled it

Rom4ik [11]3 years ago

4 0

250 gal/min
1 min = 60 sec
250gal/60sec= 4.16 gal/sec

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3 years ago

What is the least value of x which the inequality is true? 16 is greater than or equal to -2x

tia_tia [17]

Answer:Then the system of inequalities that represents the graph is second option:

2x +y is less than or equal to 3, and y-x is greater than or equal to 2

Step-by-step explanation: The area shaded is below the blue line and above the red line.

1.) The blue line goes through the points:

P1=(0,3)=(x1,y1)→x1=0, y1=3

P2=(1,1)=(x2,y2)→x2=1, y2=1

Slope: m=(y2-y1)/(x2-x1)

m=(1-3)/(1-0)→m=(-2)/(1)→m=-2

Point-slope form:

y-y1=m(x-x1)

y-3=-2(x-0)→y-3=-2(x)→y-3=-2x→y-3+3=-2x+3→y=-2x+3

The area below the blue line is:

y<=-2x+3→y+2x<=-2x+3→2x+y<=3

2x +y is less than or equal to 3

2.) The red line goes through the points:

P1=(0,2)=(x1,y1)→x1=0, y1=2

P2=(-2,0)=(x2,y2)→x2=-2, y2=0

Slope: m=(y2-y1)/(x2-x1)

m=(0-2)/(-2-0)→m=(-2)/(-2)→m=1

Point-slope form:

y-y1=m(x-x1)

y-2=1(x-0)→y-2=1(x)→y-2=x→y-2+2=x+2→y=x+2

The area above the red line is:

y>=x+2→y-x>=x+2-x→y-x>=2

y-x is greater than or equal to 2

Then the system of inequalities that represents the graph is:

2x +y is less than or equal to 3, and y-x is greater than or equal to 2

Read more on Brainly.com - brainly.com/question/11890005#readmore

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3 years ago

Divide the product of 8p and 17 by 3q.<br> pls answer

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$\frac{8p*17}{3q}= \frac{136p}{3q}\\$

136 isn't divisible by 3

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3 years ago

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2 years ago

Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell

Tanzania [10]

Answer:

Length (parallel to the x-axis): $2 \sqrt{2}$ ;

Height (parallel to the y-axis): $4\sqrt{2}$ .

Step-by-step explanation:

Let the top-right vertice of this rectangle $(x,y)$ . $x, y >0$ . The opposite vertice will be at $(-x, -y)$ . The length the rectangle will be $2x$ while its height will be $2y$ .

Function that needs to be maximized: $f(x, y) = (2x)(2y) = 4xy$ .

The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.

For Lagrange's Multipliers to work, the constraint shall be in the form: $g(x, y) =k$ . In this case

$\displaystyle g(x, y) = \frac{x^{2}}{4} + \frac{y^{2}}{16}$ .

Start by finding the first derivatives of $f(x, y)$ and $g(x, y)$ with respect to x and y, respectively:

$f_x = y$ ,
$f_y = x$ .

$\displaystyle g_x = \frac{x}{2}$ ,
$\displaystyle g_y = \frac{y}{8}$ .

This method asks for a non-zero constant, $\lambda$ , to satisfy the equations:

$f_x = \lambda g_x$ , and

$f_y = \lambda g_y$ .

(Note that this method still applies even if there are more than two variables.)

That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:

$g(x, y) = k$ .

$\displaystyle \left\{ \begin{aligned} &y = \frac{\lambda x}{2} && (a)\\ &x = \frac{\lambda y}{8} && (b)\\ & \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1 && (c)\end{aligned}\right.$ .

Replace the $y$ in equation $(b)$ with the right-hand side of equation $(b)$ .

$\displaystyle x = \lambda \frac{\lambda \cdot \dfrac{x}{2}}{8} = \frac{\lambda^{2} x}{16}$ .

Before dividing both sides by $x$ , make sure whether $x = 0$ .

If $x = 0$ , the area of the rectangle will equal to zero. That's likely not a solution.

If $x \neq 0$ , divide both sides by $x$ , $\lambda = \pm 4$ . Hence by equation $(b)$ , $y = 2x$ . Replace the $y$ in equation $(c)$ with this expression to obtain (given that $x, y >0$ ) $x = \sqrt{2}$ . Hence $y = 2x = 2\sqrt{2}$ . The length of the rectangle will be $2x = 2\sqrt{2}$ while the height will be $2y = 4\sqrt{2}$ . If there's more than one possible solutions, evaluate the function that needs to be maximized at each point. Choose the point that gives the maximum value.

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3 years ago