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yaroslaw [1]
2 years ago
12

A Gardener has 2 tulip bulb, 45 tomato plants, 108 rose bushes, and 126 herd seedling to plant in the city garden. He want each

row of the garden to have the same number of each kind of plant. What is the greatest number of rows that the garden
Mathematics
1 answer:
NNADVOKAT [17]2 years ago
3 0

Answer: <em> 9\ rows</em>

Step-by-step explanation:

<h3> <em>The complete exercise is:"A gardener has 27 tulip bulbs, 45 tomato plants, 108 rose bushes, and 126 herb seedlings to plant in the city garden. He wants each row of the garden to have the same number of each kind of plant. What is the greatest number of rows that the gardener can make if he uses all the plants?"</em></h3><h3 />

The first step to solve the exercise is to find the Greatest Common Factor (GCF) between 27, 45, 108 and 126.

You can follow these steps in order to find the GCF:

1. You must decompose 27, 45, 108 and 126into their prime factors:

27=3*3*3=3^3\\\\45=3*3*5=3^2*5\\\\108=2*2*3*3*3=2^2*3^3\\\\126=2*3*3*7=2*3^2*7

2. You must multiply the commons with the lowest exponents. Then:

<em>GCF=3^2\\\\GCF=9</em>

Therefore, the greatest number of rows that the gardener can make if he uses all the plants is:

9\ rows

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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 <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> 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<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (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 <em>t. </em>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 <em>P</em> and <em>Q</em> 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
. Over the next two days, Clinton Employment Agency is interviewing clients who wish to find jobs. On the first day, the agency
Vedmedyk [2.9K]

Answer:

4

Step-by-step explanation:

Given that :

Clients are interviewed in groups of 2 on the first day; meaning two persons at a time

Second day, clients are interviewed in groups of 4; meaning 4 persons at a time.

Therefore, if the same number of clients are to be interviewed on each day, the smallest number of clients that could be interviewed each day could be obtained by getting the Least Common Multiple of both numbers: 2 and 4

- - - - 2 - - - 4

2 - - - 1 - - - 2

2 - - - 1 - - - 1

Therefore, the Least common multiple is (2 * 2) = 4

Therefore, the smallest number of clients that could be interviewed each day is 4.

5 0
2 years ago
Work out (7x10^5)divide(2x10^2) give your answer in standard form (I WILL GIVE BRAINLIEST)
mariarad [96]
The answer to ur problem is 3.5x10^3
6 0
2 years ago
Lin multiplies 7/8 times a number. The product is less than 7/8. Which could be Lin’s number?
RSB [31]

Answer:

So, the number that Lin multiplied by 7/8 MUST be less than 1. This is evident, as any positive number multiplied by a number less than 1 will decrease, doesn't matter if its a negative or positive number.

Step-by-step explanation:

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2 years ago
A bowl is the shape of a hemisphere and made out of thick clay. The diameter of the outside bowl is 44.6 cm. And the diameter of
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Answer:

Step-by-step explanation:

Ive already explained this somebody else asked if you find it you will have the answer

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