Alright I have to take the ACT again this Saturday because I didn't take it with writing the last time I took it. Just found out the colleges wont take my writing score from this one and my composite score from the last one (28) and combine them. Which means they only accept what I'm going to make on this upcoming test. Lucky me hasn't had a math class since May. I've already taken all of the courses offered so there wasn't anything for me to take this year. Which means my math is beyond rusty.

Last time I scored a 27 on the math. I'm really hoping I can bump it up to at least a 30 this time. I've been going through a practice ACT book and for the most part I've been breezing through the math, though there are a few very simple problems that I just can't seem to figure out despite their simplicity. I credit that to just not having done any math in months. I know I can do it I just need to reengage my brain.

So if you oh so gracious fellow forumers could help me out on a few of these I'd be more than thankful!

Question 1:
http://i.imgur.com/dX3Y0.jpg the first one I had trouble with. Geometry was never my strong suit. SOLVED

Question 2:
The lengths of two sides of a triangle are 3.5 inches and 6 inches. Which of the following CANNOT be the length, in inches, of the third side? SOLVED

F.) 2
G.) 3
H.) 4
J.) 5
k.) 6

Question 3:

http://i.imgur.com/jMXfc.jpgSOLVED

What's an ACT? When I was in school we only had the SATs and I never took them even though it was "required".

I looked at the problem you posted and I gave to ask, who in the flying [censored] actually cares what the dimensions of the shaded area is? Are you ever really going to need to figure that out in real life?

If I had to figure it out in real life I'd use a damn ruler or tape measure.

ACT = American College Test. It's a comprehension college entrance test compared the critical thinking oriented SAT.

Isn't it 90% trigonometry? I took it three years ago, so my memory might be a little rusty. And for the problem you posed, can't you just chop a triangle off the top of the shaded region, rotate it 90 degrees, and stick it on the bottom right to make a smaller square with half the length and width of the larger one?

Off the top of my head, I believe an angle of some of the overlap is needed. I could be wrong, though...

Do what....



I read that about 15 times and have no idea what you're talking about haha

36 square units indeed. Pythagoras would be proud of you. Or who knows, I haven't heard of him in a while.

It is mostly Algebra I, II, and Geometry. There are some Trig and Pre-Calc questions, but as you are rusty in math you should probably just skip those entirely to give the rest of the tes your focus. Don't worry it is like 5 skipped problems.

And for your problem I would find the area of the square then see which answer is closest to 25% of the area. Don't want to spend tok much time on those problems.

Does this make sense?

http://i270.photobucket.com/albums/jj118/thebestzeldafan/mathy_zpsfbbc9899.jpg

Then, since the height and width reach from the sides to the center, it is 12/2 = 6 units in each direction. 6X6=36 square units in area.

Pre-Calc and Trig were the most recent math classes I've taken. I actually did better in both of those than in any other of my math classes. I'm more worried about the geometry and Algebra 2 than the Pre-Calc and Trig.

I have been enlightened. Thank ye good sir.

I was trying to cut off a completely different triangular shape haha.

I took a look at the problem.

The whole thing is in the fact that the corner in the middle of the square is 90° so no matter how you tilt it, it will always take up 1/4th of the square. So as you can see from this picture.

http://i.imgur.com/NGRmb.png

The red area is the same size as the empty light blue area. So if you took it away and filled it in where the blue area is you'd have 1/4th of the 12 square unit square.

And I just noticed Hyrule beat me to it...

That's what it looks like to me. The shaded top rt. triangle has a known side of 6. The blank trainalge space below the square's center also has the known side of six. I'd go with it and do the area math of 6 x 6. I vaguely remember something about similar triangles and all that lol

jeepers...6 new replies while I posted...oh well

Well then there will be nothing familiar.

Isn't it 6 x 6.

36 squared units.

This is one of those simple ones I know I already know I just can't remember it.

I vaguely remember something about a third side can be less than half of something or more than X% of the longest given side.

Question 2:
The lengths of two sides of a triangle are 3.5 inches and 6 inches. Which of the following CANNOT be the length, in inches, of the third side?

F.) 2
G.) 3
H.) 4
J.) 5
k.) 6

I choose 2 thinking that there's no way that 6 inches can stretch between 3.5 and 2 at any given angle.

Edit: Made a post based on an incorrect assumption.

Two ways:

1) wait for someone here to tell you.

2) pull up word, paint, or something with a grid and drawing on your computer. Heck, even pull out the old compass. Draw a 6 inch base, and make a circle with radius 3.5 from one end of the 6, and adjust to the other sizes and draw the circle at the other end. Connect the point of intersecting circles to the center of the 3.5 circle. Do the intersections make trianlges? Do any of the circles not intersect? Play around and create your own knowledge..it'll stick with you longer

I remembered the rule I was thinking of...I think.

Any two sides of a triangle when added must be equal to or greater than the third side. So 3.5 + (3,4,5,6) &--#62; 6 while 3.5 + 2 &--#60; 6.

Edit: Same deal as with the above edited post.

How can you not have a 3.5, 3, 6 triangle? Are you doing it for a right triangle?

I should not post advice when tired.

Just re-read the question again and I don't know why I thought it had to be a right triangle. This does show though that you can figure this stuff out and our advice can be misleading. I'll edit my post to keep others from being mislead.

Isn't #2 pretty easy to see by drawing a diagram? The vertices of the triangle all have to be at an angle > 0 degrees, no?

#2 is Pythagorean Theorem isn't it?

A^2 + B^2 = C^2

Not sure where to go after that...

I thought so too but the question does not assume it's a right triangle. He's already solved it though.