Explain 4 questions + 1 random question

**Masterthief1324** · February 5th 2010, 07:25 PM

The first 4 attached questions do not make sense to me. The 'correct' answers are dotted but I'd like to see the proof for it. (They are user-generated problems from the Sparknotes website so I anticipated some errors in them. I can't tell whether or not I'm wrong or the problem had wrong answer choices)

Status problem 1:Solved - Thanks!
Status problem 2:Unconfirmed: I checked for m when g(m) = x^2 + 4 and when g(m) = 2x + 4; 3 does not fit.
Status problem 3: [Same as problem 2]
Status problem 4:Unsolved

Random question:
Also, the forumula $\frac {p}{q}$ is used to solve polynomials correct? Polynomials and or Binomials?

**VonNemo19** · February 5th 2010, 08:04 PM

Originally Posted by Masterthief1324

The first 4 attached questions do not make sense to me. The 'correct' answers are dotted but I'd like to see the proof for it. (They are user-generated problems from the Sparknotes website so I anticipated some errors in them. I can't tell whether or not I'm wrong or the problem had wrong answer choices)

Status problem 1:Unsolved
Status problem 2:Unsolved
Status problem 3:Unsolved
Status problem 4:Unsolved

Random question:
Also, the forumula $\frac {p}{q}$ is used to solve polynomials correct? Polynomials and or Binomials?

For 1

$3p(3h)=3\Rightarrow{p}(3h)=1\Rightarrow\frac{3h}{2 }-5=1$

**Masterthief1324** · February 5th 2010, 08:47 PM

Originally Posted by VonNemo19

For 1

$3p(3h)=3\Rightarrow{p}(3h)=1\Rightarrow\frac{3h}{2 }-5=1$

It's asking for the value of 'h'.
My reasoning:
In order for the second equation to be true, ( $3p(3h)=3$ ), the function p(3h) has to equal to 1.

The function p(x) = $\frac {x}{2-5}$ or p(x) = $\frac {x}{-3}$ . Thus when x = 3h, the function p(x) becomes:

p(3h) = $\frac {3h}{-3}$

Since p(3h) HAS to equal to 1, 'h' equals -1 because:
1 = $\frac {3(-1) or (-3)}{-3}$

**VonNemo19** · February 5th 2010, 09:16 PM

Originally Posted by Masterthief1324

It's asking for the value of 'h'.
My reasoning:
In order for the second equation to be true, ( $3p(3h)=3$ ), the function p(3h) has to equal to 1.

The function p(x) = $\frac {x}{2-5}$ or p(x) = $\frac {x}{-3}$ . Thus when x = 3h, the function p(x) becomes:

p(3h) = $\frac {3h}{-3}$

Since p(3h) HAS to equal to 1, 'h' equals -1 because:
1 = $\frac {3(-1) or (-3)}{-3}$

So 4 is the wrong answer even though it was given as an answer choice. They asked for h and no answer choices are -1 so all answer choices are wrong! Take that sparknotes!

No.

Like I said before...

$3p(3h)=3$ means that $p(3h)=1$ .

Which means that

$\frac{3h}{2}-5=1$ .

Solve for h.

By the way, association is very important. you were given $p(x)=\frac{x}{2}-5$ , not $p(x)=\frac{x}{2-5}$ .

Oh, and four is CORRECT.

**HallsofIvy** · February 6th 2010, 01:49 AM

1, 3, 3, 3, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11. Just count!

Or:
1 occurs 1 time
3 occurs 3 times
5 occurs 5 times
7 occurs 7 times
9 occurs 9 times

so 11 occurs after 1+ 3+ 5+ 7+ 9 other numbers.

"Let the function g be defined by $g(x)= x^2+ 4$ . If p is a positive number such that $g(2m)= g(m)+ 22$ which of the following could be a value of m?"

There's an obvious typo in this problem. They must mean "If m is a positive number...".

$g(2m)= (2m)^2+ 4= 4m^2+ 4$ . g(2m)= g(m)+ 22 becomes $4m^2+ 4= (m^2+ 4)+ 22$ . Solve that equation for m.

Your other problem appears to be just a copy of that.

As for your random question, "Also, the forumula $\frac{p}{q}$ is used to solve polynomials correct? Polynomials and or Binomials?", I can make no sense out of it. First, $\frac{p}{q}$ is a not a "formula", it is just a fraction or, more generally, an "expression". Second, you don't "solve polynomials", you solve polynomial equations. Most importantly, since you have not said what "p" and "q" are, or how they are connected to the equation, I don't know how you would intend to use $\frac{p}{q}$ to solve the equation.

**Masterthief1324** · February 10th 2010, 01:49 PM

[quote=Masterthief1324;451914]It's asking for the value of 'h'.
My reasoning:
In order for the second equation to be true, ( $3p(3h)=3$ ), the function p(3h) has to equal to 1.

The function p(x) = $\frac {x}{2-5}$ or p(x) = $\frac {x}{-3}$ . Thus when x = 3h, the function p(x) becomes:

p(3h) = $\frac {3h}{-3}$

Since p(3h) HAS to equal to 1, 'h' equals -1 because:
1 = $\frac {3(-1) or (-3)}{-3}$

**Masterthief1324** · February 10th 2010, 02:22 PM

Originally Posted by HallsofIvy

1, 3, 3, 3, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11. Just count!

Or:
1 occurs 1 time
3 occurs 3 times
5 occurs 5 times
7 occurs 7 times
9 occurs 9 times

so 11 occurs after 1+ 3+ 5+ 7+ 9 other numbers.

"Let the function g be defined by $g(x)= x^2+ 4$ . If p is a positive number such that $g(2m)= g(m)+ 22$ which of the following could be a value of m?"

There's an obvious typo in this problem. They must mean "If m is a positive number...".

$g(2m)= (2m)^2+ 4= 4m^2+ 4$ . g(2m)= g(m)+ 22 becomes $4m^2+ 4= (m^2+ 4)+ 22$ . Solve that equation for m.

Your other problem appears to be just a copy of that.

As for your random question, "Also, the forumula $\frac{p}{q}$ is used to solve polynomials correct? Polynomials and or Binomials?", I can make no sense out of it. First, $\frac{p}{q}$ is a not a "formula", it is just a fraction or, more generally, an "expression". Second, you don't "solve polynomials", you solve polynomial equations. Most importantly, since you have not said what "p" and "q" are, or how they are connected to the equation, I don't know how you would intend to use $\frac{p}{q}$ to solve the equation.

For the question concerning g(2m), I solved and had a decimal answer:

$<br /> 4m^2 + 4 = (m^2 + 4) + 22<br />$

m = 2.7...; Therefore the answer is not 3 correct?

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