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kompoz [17]
3 years ago
7

A thin tube stretched across a street counts the number of pairs of wheels that pass over it. A vehicle classified as type A wit

h two axles registers two counts. A vehicle classified as type B with nine axles registers nine counts. During a 2​-hour ​period, a traffic counter registered 101 counts. How many type A vehicles and type B vehicles passed over the traffic​ counter? List all possible solutions.
Mathematics
1 answer:
aliina [53]3 years ago
7 0

Answer: All possible solutions would be

(1,41), (8,33), (15,25), (22,17), (29,9), (36,1)

Step-by-step explanation:

Let the number of vehicle of type A be 'x'.

Let the number of vehicle of type B be 'y'.

Since we have given that

Number of axles of type A = 2

Number of axles of type B = 9

After 2 hours, number of counts = 101

so, our equation becomes,

2x+9y=101

So, all possible solutions would be

(1,41), (8,33), (15,25), (22,17), (29,9), (36,1)

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Answer:

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Step-by-step explanation:

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3 years ago
Using the quadratic formula to solve 4x2 – 3x + 9 = 2x + 1, what are the values of x?
yaroslaw [1]

Answer:

The value of x is:

x=\dfrac{5\pm \sqrt{103}i}{8}

Step-by-step explanation:

we have to use the quadratic formula to solve for x.

The equation is given as:

4x^2-3x+9=2x+1

which could also be written as:

4x^2-3x+9-2x-1=0\\\\4x^2-3x-2x+9-1=0\\\\\\4x^2-5x+8=0

The quadratic formula for the quadratic equation of the type:

ax^2+bx+c=0 is given as:

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

Here we have:

a=4, b=-5 and c=8.

Hence, by the quadratic formula we have:

x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4\times 8\times 4}}{2\times 4}\\\\x=\dfrac{5\pm \sqrt{25-128}}{8}\\\\\\x=\dfrac{5\pm \sqrt{103}i}{8}

Hence, the value of x is:

x=\dfrac{5\pm \sqrt{103}i}{8}

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3 years ago
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tigry1 [53]

Answer:

The graduate will have $4,084.97 when he cashes in the CD at the end of the 36-months.

Step-by-step explanation:

This can be determined using the formula for calculating the future value (FV) formula as follows:

FV = PV * (1 + r)^n ................................. (1)

Where;

FV = Future value or the amount the graduate will have at end of 36 months = ?

PV = Present value or the total amount invested = Cash gifts from friends and relatives + Amount of 3 scholarships he received = $900 + $250 + $300 + $1,400 = $2,850

r = daily interest rate = 1% / 360 = 0.01 / 360 = 0.0000277777777777778

n = number of days = 36 months * 360 days = 12,960

Substituting the values into equation (1), we have:

FV = $2,850 * (1 + 0.0000277777777777778)^12,960

FV = $2,850 * 1.43332224806396

FV = $4,084.96840698229

Approximating to the nearest cent as required, we have:

FV = $4,084.97

Therefore, the graduate will have $4,084.97 when he cashes in the CD at the end of the 36-months.

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3 years ago
What is the solution to the system of equations
ExtremeBDS [4]

Answer:

Y = 2

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Substitution:

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