# Math Help - Need Help With All these Algebra Problems.

1. ## Need Help With All these Algebra Problems.

Sorry, I'm just stupid when it comes to math.

1) Rationalize the denominator of expression

2)A baseball player has a batting average of 0.175. What is the probability that he has exactly 3 hits in his next 7 at bats?
The probability is ________

3)Evaluate the expression and write the result in the form a+bi

4)Solve the following inequality. Write the answer in interval notation.

5)NASA launches a rocket at seconds. Its height, in meters above sea-level, as a function of time is given by
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after _______seconds
How high above sea-level does the rocket get at its peak?

The rocket peaks at _________meters above sea-level.

6)The equation has two solutions A and B where A<B and A=________ and B=_________

If you don't want to show the work on how you got it, it's fine. I understand that it's a lot of problems.

Sorry, I'm just stupid when it comes to math.

1) Rationalize the denominator of expression

$\frac{2}{{4 + \sqrt 5 }} = \frac{2}{{4 + \sqrt 5 }} \cdot \frac{{4 - \sqrt 5 }}{{4 - \sqrt 5 }} = \frac{{2(4 - \sqrt 5 )}}{{(4 + \sqrt {5)(4 - \sqrt 5 )} }} = \frac{{2(4 - \sqrt 5 )}}{{4^2 - 5}} = \frac{{2(4 - \sqrt 5 )}}{{11}}$

2)A baseball player has a batting average of 0.175. What is the probability that he has exactly 3 hits in his next 7 at bats?
The probability is ________
This is binomial probability.
$p=.175$
The probability of failure is,
$p=1-.175=.825$
Thus,
${7\choose 3}(.175)^3(.825)^4$

4)Solve the following inequality. Write the answer in interval notation.
That means,
$x-2\geq 5$
Or,
$x-2\leq -5$
Thus,
$x\geq 7$
Or,
$x\leq -3$
In interval notation,
$x=(-\infty,-3]\cup[-5,+\infty)$

6)The equation has two solutions A and B where A<B and A=________ and B=_________
$\begin{array}{l}
3x^2 + 15x + 3 = 0 \\
x_{1/2} = \frac{{ - 15 \pm \sqrt {15^2 - 4(3 \cdot 3)} }}{{2 \cdot 3}} = \frac{{ - 15 \pm \sqrt {225 - 36} }}{6} = \frac{{ - 15 \pm \sqrt {189} }}{6} = \frac{{ - 15 \pm 3\sqrt {21} }}{6} \\
B = x_{1 = } \frac{{ - 15 + 3\sqrt {21} }}{6} = \frac{{3( - 5 + \sqrt {21} )}}{{3 \cdot 2}} = \frac{{ - 5 + \sqrt {21} }}{2} \\
A = x_{2 = } \frac{{ - 15 - 3\sqrt {21} }}{6} = \frac{{3( - 5 - \sqrt {21} )}}{{3 \cdot 2}} = \frac{{ - 5 - \sqrt {21} }}{2} \\
\end{array}
$

5. [QUOTE=ThePerfectHacker;29042]This is binomial probability.
$p=.175$
The probability of failure is,
$p=1-.175=.825$
Thus,
${7\choose 3}(.175)^3(.825)^4$

I still do not get what the answer would be? What is multipleied into the 3 and 7?

3)Evaluate the expression $(8 + 5y)(-3+3i)$ and write the result in the form $a+bi$
$(8 + 5i)(-3+3i) \;=\;-24 + 24i - 15i + 15i^2 \;= \;-24 + 9i - 15 \;= \;-39 + 9i$

7. Originally Posted by ThePerfectHacker

That means,
$x-2\geq 5$
Or,
$x-2\leq -5$
Thus,
$x\geq 7$
Or,
$x\leq -3$
In interval notation,
$x=(-\infty,-3]\cup[-5,+\infty)$
There is error in solution.

TPH made typo (intentionaly so that you be confused ).
Solution is $x=(-\infty,-3]\cup[7,+\infty)$

8. Originally Posted by OReilly
There is error in solution.

TPH made typo (intentionaly so that you be confused ).
Solution is $x=(-\infty,-3]\cup[7,+\infty)$
That explains why it was saying it was wrong.

Originally Posted by ThePerfectHacker
This is binomial probability.
$p=.175$
The probability of failure is,
$p=1-.175=.825$
Thus,
${7\choose 3}(.175)^3(.825)^4$

I still do not get what the answer would be? What is multipleied into the 3 and 7?
Can someone explain this to me?

5) NASA launches a rocket.
Its height, in meters above sea-level, is given by: $h(t) \;=\;-4.9t^2 + 349t + 332$

At what time does splashdown occur?

How high above sea-level does the rocket get at its peak?

Splashdown occurs when $h(t) = 0.$

We have: . $-4.9t^2 + 349t + 332 \:=\:0$

Quadratic Formula: . $t \;=\;\frac{-349 \pm\sqrt{349^2 - 4(\text{-}4.9)(332)}}{2(\text{-}4.9)} \;\approx\;\{-8.3,\:72.2\}$

Splashdown occurs about $\boxed{72.2\text{ seconds}}$ after liftoff.

The maximum height occurs at the vertex of the parabola.

The vertex is at: . $t \:=\:\frac{-b}{2a} \:=\:\frac{-349}{2(\text{-}4.9)} \:\approx\:35.6$ seconds

Then: . $h(35.6)\;=\;-4.9(35.6)^2 + 349(35.6) + 332 \;=\;6546.336$

The rocket peaks at about $\boxed{546\text{ feet}}$

$_7C_3$ is another way to write it.
$\frac{7\cdot 6\cdot 5}{3!}=35$