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omeli [17]
3 years ago
10

Each student needs a pencil and a eraser to take a test.If pencils come in 8 in a box and erasers come 12 in a bag, what is te l

east number of boxes and bags needed for 24 students to each have a pencil and a eraser
Mathematics
1 answer:
charle [14.2K]3 years ago
5 0

The least number of boxes and bags needed are 3 and 2 respectively.

<u><em>Explanation</em></u>

Suppose, the number of boxes is x and the number of bags is y

Given that, pencils come in 8 in a box and erasers come 12 in a bag.

So, the total number of pencils in x number of boxes =8x and the total number of erasers in y number of bags =12y

As there are total 24 students and each student needs one pencil and one eraser to take a test, so...

8x=24\\ \\ x= \frac{24}{8}=3

and

12y=24\\ \\ y=\frac{24}{12}=2

So, the least number of boxes and bags needed are 3 and 2 respectively.

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Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt
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Answer:

y(x)=4a\sqrt{3}* sin(x)-3a

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}

But:

\frac{1}{tan(x)} =cot(x)

So:

\frac{dy(x)}{dx} =cot(x)*(3a+y)

Multiplying both sides by dx and dividing both sides by 3a+y:

\frac{dy}{3a+y} =cot(x)dx

Integrating both sides:

\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx

Evaluating the integrals:

log(3a+y)=log(sin(x))+C_1

Where C1 is an arbitrary constant.

Solving for y:

y(x)=-3a+e^{C_1} sin(x)

e^{C_1} =constant

So:

y(x)=C_1*sin(x)-3a

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y(\frac{\pi}{3} )=3a=C_1*sin(\frac{\pi}{3})-3a\\ 3a=C_1*\frac{\sqrt{3} }{2} -3a

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4 years ago
Marlene noticed that when a pair of parallel lines was cut by a transversal, two particular vertical angles were reflections of
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3 years ago
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Phoebe, Andy and Polly share £270.
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*Given
Money of Phoebe            - 3 times as much as Andy
Money of Andy                - 2 times as much as Polly
Total money of Phoebe,  - <span>£270
</span>    Andy and Polly

*Solution

Let
B - Phoebe's money
A - Andy's money
L - Polly's money

1. The money of the Phoebe, Andy, and Polly, when added together would total <span>£270. Thus, 
</span>
                  B + A + L  = <span>£270                     (EQUATION 1)

2. Phoebe has three times as much money as Andy and this is expressed as
  
                  B = 3A

3. Andy has twice as much money as Polly and this is expressed as

                  A = 2L</span>                           (EQUATION 2)
<span>
4. This means that Phoebe has ____ as much money as Polly, 

                 B = 3A
                 B = 3 x (2L)
                 B = 6L                            </span>(EQUATION 3)<span>

This step allows us to eliminate the variables B and A in EQUATION 1 by expressing the equation in terms of Polly's money only. 

5. Substituting B with 6L, and A with 2L, EQUATION 1 becomes, 

                  6L + 2L + L = </span><span>£270
</span>                                9L = <span>£270
</span>                                  L = <span>£30

So, Polly has </span><span>£30. 

6. Substituting L into EQUATIONS 2 and 3 would give us the values for Andy's money and Phoebe's money, respectively. 
</span>
                  A = 2L                           
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Andy has £60

                  B = 6L                         
                  B = 6(£30)
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Phoebe has £180

Therefore, Polly's money is £30, Andy's is £60, and Phoebe's is £180.
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Answer:

336

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Required

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