All categories

2 years ago

1. Formulate 500 lbs of a 21% CP ration using Soybean Meal (44% CP) and Corn (9% CP).

Mathematics

1 answer:

zmey [24]2 years ago

3 0

Answer:

Step-by-step explanation:

see sketch below

1 Formulate 500 lbs of a 21% cp ration using soybean meal( 44% cp) and corn (9% cp)

How many pounds of soybean meal will be used?

12 parts soy bean meal/ 35 parts total = 0.34 parts from soy bean meal

0.34 parts from soy bean meal X 500 lbs ration = 170 lbs soy bean meal

How many pounds of corn will be used ?

23 parts corn / 35 parts total = 0.66 parts from corn

0.66 parts from corn x 500 lbs ration = 330 lbs corn needed

You might be interested in

A function g(x) has x-intercepts at (StartFraction 1 Over 2 EndFraction, 0) and (6, 0). Which could be g(x)? g(x) = 2(x + 1)(x +

Tcecarenko [31]

Answer:

$g(x) = (x-6)(2x-1)$

Step-by-step explanation:

$g(x)$ intercepts the x-axis at these 2 points:

$(6,0) ;(1/2,0)$

⇒ 6 and 1/2 are roots ie; if you insert $x=6$ or $x=1/2$ into the equation of g(x) you will obtain a 0.

$g(6) = g(1/2) = 0$

now in order for 0 to appear we should have $x-6$

now in order for 0 to appear we should have $x-1/2$

but $x-1/2$ doesn't appear in any of these, but its multiple of 2 is there:

$2(x-1/2) = 2x-1$

Therefore the function;

$g(x) = (x-6)(2x-1)$

5 0

2 years ago

Read 2 more answers

Agent Hunt is transferring classified files from the CIA mainframe to his flash drive. The variable S models the size of the fil

amid [387]

Answer:

Step-by-step explanation:

$\bold{METHOD\ 1:}\\\\\text{Calculate the value of }S\ \text{in}\ 10s,\ 20s,\ 30s,\ ... \text{and difference if them}\\\\for\ t=10:\\\\S=5(10)+45=50+45=95\\\\for\ t=20:\\\\S=5(20)+45=100+45=145\\\\for\ t=30:\\\\S=5(30)+45=150+45=195\\\\195-145=50\\145-95=50\\\\\boxed{50\ MB\ per\ 10s}$

$\bold{METHOD\ 2:}\\\\S_1=5t+45\to \text{in t second}\\\\S_2=5(t+10)+45\to \text{10 second leter}\\\\\text{Difference}\\\\S_2-S_1=5(t+10)+45-(5t+45)=5t+50+45-5t-45=50\\\\\boxed{50MB\ per\ 10s}$

6 0

2 years ago

Read 2 more answers

Ana played 555 rounds of golf, and her lowest score was an 808080.

laiz [17]

Answer:

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon

posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y(t) be the amount of salt (in pounds) in the tank at time t (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

$\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)} \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)} \:\frac{pound}{s}$

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

$V(t)=100 + 2t$

We can then write the initial value problem:

$\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50$

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

$\frac{dy}{dx}+P(x)y =Q(x)$

where P and Q are continuous functions on a given interval.

In our case, we have that

$\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50$

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, $\mu \left( x \right)$ , using $\mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}$

$\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}$

Step 2: Multiply everything in the differential equation by $\mu \left( x \right)$ and verify that the left side becomes the product rule $\left( {\mu \left( t \right)y\left( t \right)} \right)'$ and write it as such.

$\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}$

Step 3: Integrate both sides.

$\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C$