• We have updated to the newest version of the forum. Please let us know if you see any issues. And, yes, we know the shoutbox is gone. We'll see about getting it back if they make a newer version.

Math help (basic to advanced)

Jana

New Member
Here's a few more I'm struggling with:

1) Two airplanes are leaving Chicago at the same time, one bound for New York and the other for California. The one heading to California has a tail wind and is thus going 25 MPH faster than the one headed for New York. If after one hour they are 575 miles apart, what are their speeds?

2) How many ounces of pure gold must be added to five ounces of something that is 1% gold in order that the mixture is 6% gold?

3) One number is five more than another number. Their sum is 285. What are they?

4) Find the equation for the line passing through (2,1) with slope 5/2.

5) Factor 36x^2-1

Any help is appreciated. I'm really struggling with this algebra crap. :(
 

dmention7

Hater
2) Solve for x by completing the square: x^2+6x=2
So, "completing the squares" is just a term for solving a quadratic equation (an equation of the form ax^2 + bx + c = 0).

The goal is to get from something that looks like x^2 + bx + c = 0 to something that looks like (x - h)^2 + k = 0.

Without going into too much detail, you can figure out what h and k are, as follows:

h = -b/2
k = c - (b^2)/4


So, you're starting with the equation x^2 + 6x = 2.

To get that into the proper form for using the rules I just gave, you need to subtract 2 from both sides of the equation to get x^2 + 6x -2 = 0

In your case, b is 6, and c is -2

Now solve for h and k like I just described:

h = -6/2 = -3
k = -2 - (6^2)/4 = -11

Substituting those values for h and k back into (x + h)^2 = k, you get:

(x+3)^2 -11 = 0 (this is the "completed square" form of your original equation)

Now, to solve for x, you can add 11 to each side of the equation to get (x+3)^2 = 11. Taking the square root of both sides, you are left with

x+3 = +/-√11 (The +/- comes from the fact that (-√11)^2 and (√11)^2 both equal 11)

Subtract 3 from both sides to get the final result of:
x = 3 + √11 or x = 3 - √11
 

dmention7

Hater
Here's a few more I'm struggling with:

1) Two airplanes are leaving Chicago at the same time, one bound for New York and the other for California. The one heading to California has a tail wind and is thus going 25 MPH faster than the one headed for New York. If after one hour they are 575 miles apart, what are their speeds?
This one is just a matter of interpreting the words into their mathematical equivalent. Let's start with the fact that distance (d) is equal to velocity (v) multiplied by the time (t) spent at that velocity.

d = v*t

In this case, the total distance between the planes is equal to the distance traveled by plane 1 plus the distance traveled by plane 2 since we can assume they are travelling in opposite directions. And since we know their total distance apart, we add the two distances together and write:

d = v1*t + v2*t where v1 and v2 are the velocities of the two planes.

We can do a little bit of simplification and change that to:

d = t * (v1 + v2)

Almost there! Since the problem states that one plane is going 25mph faster than the other, we know that v2 = v1 +25. Substituting that value in for v2, we get:

d = t * (v1 + v1 + 25) = t * (2 * v1 +25)

Now we know that the distance is 575 miles, and the time spent in the air was 1 hour so...

575 = 1 * (2 * v1 + 25) = 2* v1 + 25

Subtract 25 from both sides to get

550 = 2 * v1

Divide both sides by 2, and the final result is:

v1 = 225

Since we said before that v2 = v1 + 25, we know that v2 = 250.

So the answer is that the planes were flying 225mph and 250 mph.


2) How many ounces of pure gold must be added to five ounces of something that is 1% gold in order that the mixture is 6% gold?
First off, it helps to know that mathematically, 25% of x is .25x

Before we start with any numbers though, the problem can be set up as:

w1 * f1 + w2 * f2 = w3 * f3 where w is the total weight of each component and f is the fraction of that component that's gold. (If that doesn't make sense, let me know and I can explain it a little more).

So, "something" #1 is pure gold (100% gold, or 1.00 gold), but we don't know the weight. "Something" #2 is 1% gold (.01 gold) and weighs 5 ounces. "Something" #3 is 6% gold (.06 gold) but we don't know the weight.

w1 * 100% + 5 * 1% = w3 * 6% or 1*w1 + .01*5 = .06*w3 or finally w1 + .05 = .06*w3

The other piece of info we have is that w3 (the final weight in ounces of the something) is equal to w1 + 5. Substituting that in to the last equation:

w1 + .05 = .06 * (w1 + 5) or w1 + .05 = .06*w1 + .3

Subtract .06*w1 from both sides and you get

.94*w1 + .05 = .3

Subtract .05 from both sides:

.94*w1 = .25

Divide both sides by .94 and you get

w1 = .266 ounces (approximately)

3) One number is five more than another number. Their sum is 285. What are they?
If we call the first number X and the second number Y, we can write

X + Y = 285

From the problem statement, X is five more than Y, or X = Y +5, so

X + (X + 5) = 285

or

2*X + 5 = 285

Subtract 5 from both sides

2*X = 280

Divide both sides by 2

X = 140


4) Find the equation for the line passing through (2,1) with slope 5/2.
There's kind of a shortcut in the way this one's worded, but shortcuts are no good unless you know how to get there the long way ;)

The equation for a line can be written y = mx + b

where m is the slope and b is the y-intercept (the value of y when x=0)

We already have the slope (m = 5/2 = 2.5, so we just need to find b. Since we know that the line passes through the point (2,1), then these x and y values must satisfy the line's equation. In other words, just plug that point into the line's equation and solve for b

1 = 2.5 * 2 + b

or

1 = 5 + b

Subtract 5 from both sides, and

b = -4

So the equation for the line is: y = 2.5x - 4


5) Factor 36x^2-1
When it comes to factoring equations with squares, there a couple of different forms to check to see if your equation fits, because they will make your job easier. The relevant one here is called "difference of squares", and that's when you have an equation that looks like:

a^2 - b^2

If so, it factors to (a+b)*(a-b). It might not be immediately obvious, but your equation fits this form.

Since 36 = 6^2, 36x^2 is the same as (6^2)(x^2) which is the same as (6x)^2. Therefore, your "a" term is 6x.

The "b" term is pretty obvious here... 1 = 1^2, so your "b" term is 1.

So here's how it all flows:

36x^2 - 1 = (6x)^2 - 1^2 = (6x+1)*(6x-1)

The factored form therefore is (6x+1)*(6x-1)

To convince yourself, try multplying it all out and seeing if it simplifies back down to 36x^2-1




Whew! time for a :draught2:
 
Last edited:
Z

Z-licious

Guest
I re-read my book to try and understand how you know to take the coefficient thing... I just don't get that.
Completing the square means you have to get one side of your equation into (x + a)^2 form, where 'a' is a constant, not a variable. To do this, you want to first get all your variables on one side with the coefficient of your x^2 term equal to 1, and just a constant number on the other. Your equation already starts in this form, making that step easy:

x^2 + 6x = 2

Next, you have to complete the square. To do this, you take coefficient of your x term (6), and divide by 2, which equals 3. Then, you square this value, 3^2 to get 9. This is the value you need to get on the left side of your equation to "complete the square", so add 9 to both sides.

x^2 + 6x + 9 = 11

Now you can convert x^2 + 6x + 9 to (x+3)(x+3) or (x+3)^2 for short. Multiply everything out to check your work.
 

Jana

New Member
I'm struggling with a few more...

1) Find the equation for the line passing through the origin and perpendicular to y= 1/2x-2.

2) Simplify: (x^2y^11z^14)^-2

3) Solve x^2-7x-3=0 for x using any method you prefer


The other ones are graph related, so they are harder to show online. I'm going to see if a co-worker or my boss can look them over tomorrow.

Thanks for helping me!

Jana
 

Big Nate

Chaos Engineer
1) Well this one you know the slope is opposite of the line you have be cause it tells you it it perpendicular. The origin is (0,0) so you now have the slope and a point on the line to plug into the equation and run it through.

M=-2 (opposite inverse of the slope of the line you have)
y=0 (y-intercept has to be zero because the line passes through the (0,0))
x=0 (x-intercept)
y=-2x+0

2) Here we can apply the power rule. This means we can take the exponent from each verifiable and subtract it from the exponent from the outside of the parenthesis.

first we can rewrite it as this.
(x^2)^-2(y^11)^-2(z^14)^-2
then subtract the exponents. THIS IS WRONG
x^2-(-2) y^11-(-2) z^14-(-2)
so we end up with.
x^4y^13z^16

3) This is a quadratic which means you can use the quadratic equation. http://upload.wikimedia.org/math/3/e/a/3ea647783b5121989cd87ca3bb558916.png

or you can factor it (but yours is prime so you can't factor it)

So lets plug in the numbers and run it through the quadratic equation.

-(-7)±√ (-7)^2-4(1)(-3)
/
2(1)
solve the square first and get
7±√61
/
2
That is the exact answer.


guys check my stuff
Jana let the guys confirm this before you take it as good.
 

Jana

New Member
1)

2) Here we can apply the power rule. This means we can take the exponent from each verifiable and subtract it from the exponent from the outside of the parenthesis.

first we can rewrite it as this.
(x^2)^-2(y^11)^-2(z^14)^-2
then subtract the exponents.
x^2-(-2) y^11-(-2) z^14-(-2)
so we end up with.
x^4y^13z^16
.

I gave you the wrong number for the first set....but also do not understand where you came up with your numbers....

(x^-7 y^11 z^14)^-2

Can someone help me with this? It's a questions that's due today....

Thanks!
 

Jana

New Member
3) This is a quadratic which means you can use the quadratic equation. http://upload.wikimedia.org/math/3/e/a/3ea647783b5121989cd87ca3bb558916.png

or you can factor it (but yours is prime so you can't factor it)

So lets plug in the numbers and run it through the quadratic equation.

-(-7)±√ (-7)^2-4(1)(-3)
/
2(1)
solve the square first and get
7±√61
/
2
That is the exact answer.


guys check my stuff
Jana let the guys confirm this before you take it as good.
Can someone (or Nate) walk through the steps a little clearer? I don't understand how that flows to the answer of 2.
 

Workdawg

NARWHAL
The exact answer is actually (7±√61)/2

not just 2.

It's kind of unclear because he put the denominator on the line below.

To clarify a bit though... here is the quadratic equation:


a, b, and c are the numbers in front of the x in the starting equation.

x^2-7x-3=0

a is in front of x² = 1
b is in front of the x = -7
c is in front of x^0 (essentially the regular number) = -3

So you just have to plug a b c into the quadratic equation, and you get

-(-7)±√ (-7)^2-4(1)(-3)
/
2(1)

 
Last edited:

Workdawg

NARWHAL
I missed your first reply...

(x^-7 y^11 z^14)^-2


The first step is to distribute the -² to each variable.

(x^-7)-² (y^11)-² (z^14)-²

Then you just MULTIPLY (I assumed Nate was correct on the adding them, but he just pointed out it's supposed to be multiplied) the exponents together. You could think of it like this I suppose...

x^(-7*-2) y^(11*-2) z^(14*-2)

Then just do that math and you end up with:

(x^14)(y^-22)(z^-28)
 
Last edited:

Big Nate

Chaos Engineer
Trying to put some of this math into a way that is understandable is tough on the forum. But Dan cleared up what i was trying to say.
 

Big Nate

Chaos Engineer
Here is a good thing to know for exponents. As i read my post i did it wrong i should have multiplied the exponents instead of add them.

(b^M)^N = b^m*n
(5^2)^3 = 5^6
 

Jana

New Member
Thank you both! Yes, online is hard (and it's an online class). Jess is actually going to start to tutor me, so I'm hoping that will help :D
 
Top