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andreev551 [17]
3 years ago
7

Please help me with this math word problem

Mathematics
1 answer:
hjlf3 years ago
3 0
Answer: The cost of 16 pencils and 10 notebooks is $5.84.


Explanation: You can solve this using linear systems:

Let X be the cost per pencil, and let Y be the cost per notebook.

(1) 7x+8y=4.15
(2) 5x+3y=1.77

Choose a variable to eliminate. I’ll eliminate X first as an example. To eliminate a variable, you must have the same coefficient beside the variable for both equations.

Equation 1 now becomes:

(3) 35x+40y=20.75

Equation 2 now becomes:

(4) 35x+21y=12.39

Now that you have 35 as a coefficient for X in both equations, you can subtract the two equations to officially eliminate it!

(3)-(4)

19y=8.36
y=0.44

Now that you have the value of Y, substitute that value into either one of the equations to get X.

Substitute y=0.44 in (1)

7x+8(0.44)=4.15
7x+3.52=4.15
7x=0.63
x=0.09

Therefore, the cost per pencil is $0.09/pencil and the cost per notebook is $0.44/notebook.

Almost done lol...

To find the cost of 16 pencils, multiply 0.09*16. That gives you $1.44, which will be the cost of 16 pencils.
The cost of 10 notebooks is 0.44*10, which gives you $4.40. That’s the cost of 10 notebooks.

To find the total price, add these values together!

1.44+4.40=5.84

Therefore, the cost of 16 pencils and 10 notebooks is $5.84.

Hope that helps シ
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Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

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The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.  

Remember, 1 h = 60 minutes. The initial value problem is  

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x =  - 6* 10⁻⁴*(3x - 10⁻¹)               x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us  

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C    or    |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).  

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Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then  

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If  t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

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Answer:

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Step-by-step explanation:

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