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1 year ago

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A group of friends wants to go to the amusement park. They have no more than $320 to spend on parking and admission. Parking is

$9.25, and tickets cost $28.25 per person, including tax. Write and solve an inequality which can be used to determine p, the number of people who can go to the amusement park.

1 answer:

Hunter-Best [27]1 year ago

8 0

A group of friends wants to go to the amusement park. They have no more than $320 to spend on parking and admission. Parking is $9.25, and tickets cost $28.25 per person, including tax. Write and solve an inequality which can be used to determine p, the number of people who can go to the amusement park.

___________________________________________

Please, give me some minutes to take over your question

________________________

They have no more than $320 to spend on parking and admission

320 ≥ 9.25 + 28.25*p

9.25 + 28.25*p ≤ 320

320- 9.25 ≥ 28.25*p

310.75/28.25 ≥ p

p ≤ 11

__________________________

Answer

The number of people who can go to the amusement park can be a maximum of 11 people (less than or equal to 11).

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3 years ago

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Answer:

m=7

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If q=1 then r=3 or r=-1.

If q=2 then r=3.

They are probably looking for q=1 and r=3 because the other combinations were used earlier in the problem.

Step-by-step explanation:

Let's assume the remainders left when doing P divided by (x-1) and P divided by (2x+3) is R.

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$=-\frac{27}{4}+\frac{9m}{4}-5$

$=\frac{-27+9m-20}{4}$

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Both of these are equal to R.

$m-3=R$

$\frac{9m-47}{4}=R$

I'm going to substitute second R which is (9m-47)/4 in place of first R.

$m-3=\frac{9m-47}{4}$

Multiply both sides by 4:

$4(m-3)=9m-47$

Distribute:

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$-12+47=5m$

Simplify left hand side:

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Divide both sides by 5:

$\frac{35}{5}=m$

$7=m$

So the value for m is 7.

$P(x)=2x^3+7x^2-5$

What is the remainder when dividing P by (x-1) or (2x+3)?

Well recall that we said m-3=R which means r=m-3=7-3=4.

So the remainder is 4 when dividing P by (x-1) or (2x+3).

Now P divided by (qx+r) will also give the same remainder R=4.

So by remainder theorem we have that P(-r/q)=4.

Let's plug this in:

$P(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-r}{q})^2-5$

Let x=-r/q

This is equal to 4 so we have this equation:

$2u^3+7u^2-5=4$

Subtract 4 on both sides:

$2u^3+7u^2-9=0$

I see one obvious solution of 1.

I seen this because I see 2+7-9 is 0.

u=1 would do that.

Let's see if we can find any other real solutions.

Dividing:

1 | 2 7 0 -9

| 2 9 9

-----------------------

2 9 9 0

This gives us the quadratic equation to solve:

$2x^2+9x+9=0$

Compare this to $ax^2+bx+c=0$

$a=2$

$b=9$

$c=9$

Since the coefficient of $x^2$ is not 1, we have to find two numbers that multiply to be $ac$ and add up to be $b$ .

Those numbers are 6 and 3 because $6(3)=18=ac$ while $6+3=9=b$ .

So we are going to replace $bx$ or $9x$ with $6x+3x$ then factor by grouping:

$2x^2+6x+3x+9=0$

$(2x^2+6x)+(3x+9)=0$

$2x(x+3)+3(x+3)=0$

$(x+3)(2x+3)=0$

This means x+3=0 or 2x+3=0.

We need to solve both of these:

x+3=0

Subtract 3 on both sides:

x=-3

----

2x+3=0

Subtract 3 on both sides:

2x=-3

Divide both sides by 2:

x=-3/2

So the solutions to P(x)=4:

$x \in \{-3,\frac{-3}{2},1\}$

If x=-3 is a solution then (x+3) is a factor that you can divide P by to get remainder 4.

If x=-3/2 is a solution then (2x+3) is a factor that you can divide P by to get remainder 4.

If x=1 is a solution then (x-1) is a factor that you can divide P by to get remainder 4.

Compare (qx+r) to (x+3); we see one possibility for (q,r)=(1,3).

Compare (qx+r) to (2x+3); we see another possibility is (q,r)=(2,3).

Compare (qx+r) to (x-1); we see another possibility is (q,r)=(1,-1).

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3 years ago