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2 years ago

= In a 12-egg carton, 1/6 equals 2 eggs. Use the grids below to help you

Mathematics

1 answer:

anzhelika [568]2 years ago

8 0

Answer:

given below

Step-by-step explanation:

These are fractions problems .

If you have 12 eggs and you need 1/6 then

12* 1/6= 12/6= 2 eggs

Similarly if you need 1/3,1/4, 1/2

12 * 1/3= 12/3= 4 eggs

12 * 1/4 = 12/4= 3 eggs

12 * 1/2 = 6 eggs

These can be drawn in sets of 12.

from the picture we see that dividing the 12 eggs in two parts and coloring one part gives 6 eggs.

Similarly dividng the 12 eggs in three parts and coloring one part gives 4 eggs.

The fractions 1/3,1/4,1/2 or 1/6 means dividing the 12 eggs into 3,4,2, or 6 parts and obtaining 1 part out of it.

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Jason wanted to place all 540 marbles in the 3 newly bought boxes. If he wants to

kotegsom [21]

Answer:

The first box will have 90 marbles, the second 180 and the third 270.

Step-by-step explanation:

Division in a ratio of 1:2:3

1 + 2 + 3 = 6

The first box will have 1/6 of the marbles.

The second box will have 2/6 = 1/3 of the marbles

The third box will have 3/6 = 1/2 of the marbles.

First box:

One sixth, so:

(1/6)*540 = 540/6 = 90

Second box:

One third, so:

(1/3)*540 = 540/3 = 180

Third box:

One half, so:

(1/2)*540 = 540/2 = 270

The first box will have 90 marbles, the second 180 and the third 270.

7 0

2 years ago

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.

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

Step 4: Find the value of the constant and solve for the solution y(t).

$50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000$

$y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}$

Now, the tank is full of brine when:

$V(t) = 400\\100+2t=400\\t=150$

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

$y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75$

6 0

2 years ago

Last year,Mark was 46 inches tall.This year,Marks height is 3 inches less than peters height.Peter is 51 inches tall.How tall is

In-s [12.5K]

Im not sure abbout the equations... but Mark is 48 inches, cuz it says; He WAS 46 in. tall. this year, amrks height is 3 inches LESS than peters height. So now, peter is 51 in. tall How tall is mark NOW? So u just have to minus 51-3 = 48!

7 0

2 years ago

Read 2 more answers

The increase in a person's body temperature to, above 98.6°F, can be modeled by the function (O)- 2 1. where !

elixir [45]

Horizontal asymptote is what happens to T(f) as t becomes extremely large (approaches infinity). The horizontal asymptote is the line which T(t) approaches as t goes to infinity, i.e. the line y=0. This means that the person's body temperature will approach 0ºF above 98.6ºF as t goes to infinity. i.e. the person's body temperature will approach 98.6ºF as time lapses.

The first option.

-Hops

6 0

2 years ago

La distancia de Urano al Sol es 2 870 990 000 km y la distancia de la Tierra al Sol es 1,496 × 108 km. Aproximadamente, ¿cuántas

Leni [432]

Answer:

2,7204x 10^9

Step-by-step explanati

2870000000-

149600000

6 0

2 years ago