Any help is really appreciated...

2. Originally Posted by Paulo1913

Any help is really appreciated...
It would be easier if you posted your work and then we could comment on it
directly.

RonL

3. Hello, Paulo1913!

It's hard to see your work from here,
. . but it looks like you played the $\displaystyle 7\heartsuit$ instead of $\displaystyle K\spadesuit$,
. . you transposed to the key of $\displaystyle G$ major instead of $\displaystyle E\flat$ minor,

4. I can't post my answers here as they would take forever to write out... even if you could do one or two questions it would be appreciated...

I really need help with this...

6. Oh, you can't be serious. Just post your final answer. Whatever it might be. 204? 39.5? 1958357? It'll only take a minute.

7. 1.a 1.961505..

1.b 0.718242..

RonL

8. For question2.a: use the Pythagorean theorem, $\displaystyle a^{2}+b^{2}=c^{2}$
where a= length of triangle
b = height of triangle
c = hypotenuse of triangle

For example, if you had a triangle with a length of 3, and a height of 4 find the hypotenuse:

1. $\displaystyle 3^{2}+4^{2}=c^{2}$

2. $\displaystyle 9+16=c^{2}$

2. $\displaystyle 25=c^{2}$

2. $\displaystyle \sqrt{25}=c$

2. $\displaystyle 5=c$

So the hypotenuse would be 5.

So figure out what you know (hint: you know the length and the height) and solve for what you want to know (the paper will tell you what you want to know)

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For 2.b
The rectangle will have the same length on it's top side as it has on it's bottom side, so the top of the box is 2.8m long. The triangle shares this side, so the triangle has a length of 2.8 and a height of 1.2

You should know that tangent of an angle tells you the opposite side divided by the adjacent side (in this case, height divided by width).

So $\displaystyle tan(EHD)=\frac{ED}{FG}$

Plug in the values you know:
$\displaystyle tan(EHD)=\frac{1.2}{2.8}$

Solve for EHD by taking the arctangent of both sides:
$\displaystyle EHD=arctan(\frac{1.2}{2.8})$

Now just plug it into your calculator:
$\displaystyle EHD\approx 23.2$º

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2.c
Look at the problem and see that if you draw a horizontal line straight left from point K, it will intersect line EF, label that point "P" Now angle PKH is a 90º angle. This means you only have to solve for PKF, then add 90.

Now PKF will be the same as KFG, (I can explain that if you need, or you can think about it for a while, remembering that there are 180 degrees in a triangle, and 90 degrees at the corner of a rectangle)

So we need to solve for KFG, and add 90. Why don't you try, using the same technique from 2.b (but this time it won't be tangent, think about what values you know, then decide which trigonometric function you need to use)

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2.d
Look at the problem and see that CB=JI

Since you have AIand JI, you can use the Pythagorean theorem to find AJ (as we did in question 2.1)

Now, with AJ and DA, you can use the Pythagorean theorem to find DJ, I already told you how to find AJ, can you figure out how to find DA? (hint, GC=HJ)