Marine Study Guide Math ?'s

[EA-6B1]

Questions from the Math section of the Marine SG...

#28, about the three company grade officers. Can someone please explain?

#34, just need the equation as to how you come up with the answer.

#41, do you just use the answers to solve for 'X' amount of balls?
3x+x=(the answer provided)

#42, this is the distance formula, right?

#47, I know area of a triangle is (1/2)base*height... but not for this one. What do you do for a triangle that has all sides the same?

Thank you all. I really appreciate the help.

 

[Eteled]

I can’t fall asleep so what the heck.

#28, three variable linear algebra, but its complexity doesn’t warrant a formal setup.
To make things seem less complicated lets rename and solve. 1stLt = X, 2ndLT = Y Captains = Z. Setup the three equations given in the question.

X + Y = 30, Y + Z = 45, X + Z = 35

Now solve one equation for whatever variable you desire.

X = 30 – Y Plug this sucker into your other equation containing X

(30 – Y) + Z = 35 Solve this sucker for either Y or Z, to be fast pick Z. That way you don’t have to mess with minus signs

Z = 5 + Y Now stick this into the other equation containing Z.

Y + (5 + Y) = 45 Solve for Y. Y = 20 One down! The next are easy. Plug your Y value in for both equations containing it.


X + 20 = 30, X = 10 and Z + 20 = 45, Z = 25

So the total is X + Y + Z = 10 + 20 + 25 = 55 You could pick any unknown at the start and reached the same answer. The steps would differ, but the procedure is the same.


#34, this is a “trick” question. X*Y = Y(X+1) Ignore math logic! Clearly this equation isn’t true, but since they said it is, then it IS.

So you’re given –4*J = -6. You can’t rely on traditional math logic. The trick is to list what information has been given to you.

Specifically, X = -4 and the hard one… Y(X+1) = -6. Now simply put in your X value…

Y(-4+1) = -6 which becomes –3Y = -6 or Y = 2.


#41, yeah so you’ve got 3X + X = 4X. So X must be a multiple of 4. Only one choice is a multiple of 4.

#42, just use that old a^2 + b^2 = c^2 logic.

#47, 1/2b*h does work here. You just need to determine h. I’m sure there’s a fancy formula to solve area for an equilateral triangle, but I don’t know it. I solved this question by chopping the triangle into two identical right triangles (base of 3, hypotenuse of 6). Then apply pythagoras theorem jes like #42. Of course you’ll have to rewrite it to solve for the missing “leg” instead of the hypotenuse. And then you’ll have h and you can solve for either your half sized triangles or the big one.

 

[EA-6B1]

First off, Thanks Eteled for working those out for me...

Ok, on #42, I don't see how I can use the Pythagoreom Theorem to find the length of XY, given the coordinates?

And on #47, when I cut the big triangle into two right triangles, I get a base of 3, a hypotenuse of 6, and the other leg (or height of triangle) as 5. Now, when I take the Area of the big triangle: A=(1/2)6*5... I get 15 for the Area. Obviously, I'm missing something. Any help?

 

[Eteled]

#42, you need to subtract the coordinates to determine the distance. So we’ve got points (1,2) and (4,6), which tells us that they are a 3 x and 4 y apart. Now these numbers correspond to the amount of x and y displacement independently. You want the shortest distance between the points. Imagine drawing the x and y components and what do you get, a triangle with sides 3 and 4 and the shortest distance being the hypotenuse. Hence you use Pythagoras to determine the distance. Now this is all visual and it’ll make sense if you draw in the x and y distances by hand. Once you get used to this you should be able to solve it without drawing a picture ((x2-x1)^2+(y2-y1)^2)^(1/2) = distance. Where x1, x2, y1, y2 are the coordinates to each point. Don’t worry about abs here because the ^2 will take care of any sign trouble for you.

#47, so lets rewrite Pythagoras first, a^2 + b^2 = c^2 … b = (c^2 – a^2)^(1/2). Ok now chopping the triangle into two we get a = 3, c = 6 and plugging this into our equation for b we get b = (36 – 9)^(1/2) = sqrt(27) . You might have written sqrt(25) = 5. Just something you slipped on. They clean sqrt(27) up. 27 = 9*3, so they write sqrt(9*3) and then they pull out the 9. sqrt(27) = 3*sqrt(3). Now you have h. Put it into ½*b*h and you get ½*6*3*sqrt(3) = 9sqrt(3).

 

[EA-6B1]

You're the Man! My bad on the subtraction of 36-9=25? I'm too calculator dependent. THanks a lot man.

 

[BigSkyGuy]

Originally posted by Eteled
I can’t fall asleep so what the heck.

#28, three variable linear algebra, but its complexity doesn’t warrant a formal setup.
To make things seem less complicated lets rename and solve. 1stLt = X, 2ndLT = Y Captains = Z. Setup the three equations given in the question.

X + Y = 30, Y + Z = 45, X + Z = 35

Now solve one equation for whatever variable you desire.

X = 30 – Y Plug this sucker into your other equation containing X

(30 – Y) + Z = 35 Solve this sucker for either Y or Z, to be fast pick Z. That way you don’t have to mess with minus signs

Z = 5 + Y Now stick this into the other equation containing Z.

Y + (5 + Y) = 45 Solve for Y. Y = 20 One down! The next are easy. Plug your Y value in for both equations containing it.


X + 20 = 30, X = 10 and Z + 20 = 45, Z = 25

So the total is X + Y + Z = 10 + 20 + 25 = 55 You could pick any unknown at the start and reached the same answer. The steps would differ, but the procedure is the same.

Also, you can solve this way:

x + y = 30
y + z = 45
x + z = 35

Total your three lines up:
2x + 2y + 2z = 110

Divide by 2:
x + y + z = 55

A little faster and proof that often times there is more than one way to skin the cat.

 

[EA-6B1]

Nice. Thanks, BigSkyGuy. I have a question on the Nautical Section:

#35 It says to define Aspect Ratio and tell what the difference is in High and Low aspect Ratio. On the answer sheet, it defines it as the ratio of the distance between the wing tips of an airplane to its average wing width. Can someone break this down into laymen's terms for me?

One more. What about Yaw? I know it's the rotation of an aircraft around it's Y-axis, but what else are we supposed to know?

I'm sure there's more, but that's all for now. Thanks.

 

[Eteled]

Bigskyguy's method is a traditional linear algebra approach. I didn't want to write such a setup because it could be confusing to someone that hasn't seen such math before.

 

[EA-6B1]

Bump... any help?

 

[Alex Fowler]

On #32 (the x star y question), the star doesn't represent multiplication--it represents a completely different, made-up function.

 

[Eteled]

It doesn't matter what * is. You don't use it to solve the question. Listing the given information is the key to this type of question.

 

[EA-6B1]

What about these questions guys?
#35 It says to define Aspect Ratio and tell what the difference is in High and Low aspect Ratio. On the answer sheet, it defines it as the ratio of the distance between the wing tips of an airplane to its average wing width. Can someone break this down into laymen's terms for me?

One more. What about Yaw? I know it's the rotation of an aircraft around it's Y-axis, but what else are we supposed to know?

I'm sure there's more, but that's all for now. Thanks.

 

[EA-6B1]

More Math ?'s too...

#20 talking about the airfields, how do you find their lengths?

#44. I just need the easiest logic on this one. I figured it out, but it took a while.

Sorry for all the questions. Thanks a lot

 

[deejaymo]

? 42 there are a group of right triangles we all have learned, I think this question was best solved by recognition of the legs 3 and 4. Anytime your see a right triangle with the legs 3 and 4 the hyp.= 5

 

[deejaymo]

?20 Width x Length =area; here they give you the widths of each field so

(100 x X) + (200 x X) = 360,000

100x + 200x = 360,000

300x = 360,000

x= 1200