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Bas_tet [7]
1 year ago
8

a child starts a piggy bank with $2. each day, the child receives 25 cents at the end of the day and puts it in the bank. if A r

epresents the amount of money and d represents for the number of days then A(d)=2+0.25d gives the amount of money in the bank as a function of days. a)explain why the domain does not contain the value d=2.5. b)explain why the range does not include the value A=$3.10
Mathematics
1 answer:
Troyanec [42]1 year ago
8 0

For a given function f(x), the domain is the set of possible values of x that we can use and the range is the set of the possible values of f(x).

For both points we will find that the answer is:

"We only can evaluate the function with integer values of d."

a) Here we have the function:

A(d) = 2 + 0.25*d

The first thing we want to answer is:

"explain why the domain does not contain the value d=2.5 "

We know that at the end of each day the kids get's the $0.25, so at d= 2.5 there two days and a half, and because only 2 nights passed, at this moment the kid only got the money two times.

So for d = 2.1

           d = 2.5

           d = 2.7

etc

We would get the exact same total money, thus we only can evaluate the function with integer values of d.

b) Why the range does not include the value A = $3.10?

Remember that d can only be an integer number.

Now let's solve:

A(d) = 3.10 = 2 + 0.25*d

          3.10 - 2 = 0.25*d

          1.10 = 0.25*d

           1.10/0.25 = d = 4.4

So we only can get  A = $3.10 if d = 4.4, and remember that d only can be a whole number, this is why the range does not include the value $3.10

If you want to learn more, you can read:

brainly.com/question/1632425

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Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

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The solution is "The maximum number of solutions is one."

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The equation for the circle given as we know from the problem without further analysis is so far x^2+y^2=r^2.

The equation for the parabola without further analysis is y=ax^2+9.

We are going to plug ax^2+9 into x^2+y^2=r^2 for y.

x^2+y^2=r^2

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Subtract r^2 on both sides.

x^2+a^2x^4+18ax^2+81-r^2=0

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a^2x^4+(18a+1)x^2+(81-r^2)=0

The discriminant of the left hand side will tell us how many solutions we will have to the equation in terms of x^2.

The discriminant is B^2-4AC.

If you compare our equation to Au^2+Bu+C, you should determine A=a^2

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The discriminant is

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Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

(u+v)^2=u^2+2uv+v^2

(324a^2+36a+1)-4a^2(81-r^2)

Distribute the 4a^2 to the terms in the ( ) next to it:

324a^2+36a+1-324a^2+4a^2r^2

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We know that r>0 because in order it to be a circle a radius has to exist.

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That is,

x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}

means you have:

x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}

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x=\pm \sqrt{\frac{-(18a+1)-\sqrt{36a+1+4a^2r^2}}{2a^2}}.

The second one is definitely includes a negative result in the square root.

18a+1 is positive since a is positive so -(18a+1) is negative

2a^2 is positive (a is not 0).

So you have (negative number-positive number)/positive which is a negative since the top is negative and you are dividing by a positive.

We have confirmed are max of one solution algebraically. (It is definitely not 3 solutions.)

If r=9, then there is one solution.

If r>9, then there is two solutions as this shows:

x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}

r=9 since our circle intersects the parabola at (0,9).

Also if (0,9) is intersection, then

0^2+9^2=r^2 which implies r=9.

Plugging in 9 for r we get:

x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2(9)^2}}{2a^2}}

x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+324a^2}}{2a^2}}

x=\pm \sqrt{\frac{-(18a+1)+\sqrt{(18a+1)^2}}{2a^2}}

x=\pm \sqrt{\frac{-(18a+1)+18a+1}{2a^2}}

x=\pm \sqrt{\frac{0}{2a^2}}

x=\pm 0

x=0

The equations intersect at x=0. Plugging into y=ax^2+9 we do get y=a(0)^2+9=9.  

After this confirmation it would be interesting to see what happens with assume algebraically the solution should be (0,9).

This means we should have got x=0.

0=\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}

A fraction is only 0 when it's top is 0.

0=-(18a+1)+\sqrt{36a+1+4a^2r^2}

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18a+1=\sqrt{36a+1+4a^2r^2

Square both sides:

324a^2+36a+1=36a+1+4a^2r^2

Subtract 36a and 1 on both sides:

324a^2=4a^2r^2

Divide both sides by 4a^2:

81=r^2

Square root both sides:

9=r

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