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#1 Help Me ! » A question that has both my parents stumped! Quadratics. » 2022-06-21 20:52:46

simonmagusflies
Replies: 1

(ax + 3)(5x^2 - bx + 4) = 20x^3 - 9x^2 - 2x + 12
The equation above is true for all x, where a and b are constants. What is the value of ab?
A) 18
B) 20
C) 24
D) 40

The answer is C, 24. For the first term, you multiply (ax + 3) into the second bracket term to get 5ax^3 + 15x^2 - abx^2 - 3bx + 4ax + 12 = 9x^2 - 2x + 12. That much I got. But, the next step is to eliminate all numbers with non-x^2 terms?? So you get 15x^2 - abx&^2 = 9x^2. THAT I don't get. Neither do my parents. Can someone please explain what reasoning they used to eliminate all numbers with non-x^2 terms?

Thank you!

#2 Help Me ! » Help me with 2 trig questions, please » 2021-10-17 23:19:37

simonmagusflies
Replies: 1

1. I know this one is relatively easy, I just cannot for the life of me remember these tricky angle rules, so please help.
In the diagram shown, parallel lines m and n are cut by transversal p.
https://i.imgur.com/ScllcCF.png
Which statement about angle 1 and angle 2 must be true?
A) Angle 1 ≅ Angle 2
B) Angle 1 is the complement of Angle 2
C) Angle 1 is the supplement of Angle 2
D) Angle 1 and Angle 2 are vertical angles

2. What is the value of x in this diagram?
https://i.imgur.com/Mpokyjc.png
A) 60 deg
B) 72 deg
C) 90 deg
D) 120 deg

Thanks!

#3 Re: Help Me ! » Area of polygons » 2021-08-31 19:51:50

Hello, I think I've got it, but I'm not sure if it's correct. If you don't mind, can you please check them for me?

#1. Solve the triangle with the Pythagorean theorem: a^2 + 12^2 = 13^2, a = 5. Find the area of the triangle: (5 x 12)/2 = 30. Multiply 30 by 2 because there are two triangles. The area of the rectangle is 60.

#2. Use the Pythagorean theorem to solve the triangle: 12^2 + b^2 = 15^2, get 9. Solve for the area: (12 x 9)/2 = 54. The rectangle's missing side length is 19 because the triangle's side is 9, 9 + 10 = 19, use to solve for area. 8 x 19 = 152. Add together to get 206, which is the area of the whole polygon.

Thanks!

#4 Help Me ! » Area of polygons » 2021-08-31 17:54:19

simonmagusflies
Replies: 4

Hello, I gotta find the areas of these two shapes, but I'm having a ton of trouble. I just can't figure out where to start. Can someone give me some pointers, please? Thanks.

#1: https://i.imgur.com/4PIIUjw.png
#2: https://i.imgur.com/rdP6UPM.png

#5 Help Me ! » This circles question set that is the bane of my existence right now » 2021-07-27 20:16:26

simonmagusflies
Replies: 1

I'm having a lot of trouble with this question -- my mom completely lost her cool with me because I just couldn't get my head around it! So now I'm on my own until I can solve it, which I can't, because I'm very much trash at math.

A farmer ties a horse to a building on a 50 foot lead.   The building measures 20 feet by 20 feet (floor).  What is the maximum area the horse can use for grazing? If there are regions you can't find the area of, provide as good an estimate as you can. Assume the horse is tied to a corner outside the building, cannot get in, and that the building is not grazing area. (Remember, this will be based on parts of circles, no other shapes...the horse's rope will only get shorter when he tries to go around the building...)

https://i.imgur.com/w0x2mU4.png

1. How much of the 50-foot circle can the horse reach without getting interrupted by the building? What is that area?
A: So, since the radius (the lead) is 50, the area of the whole circle is 7853.98? And since it's only the 50-foot circle, I gotta remove the quarter part, 7853.98/4 = 1963.495, so 7853.98 - 1963.495??? I'm dying. 5890.485. Is that it? Surely not.

2. Assume the horse has grazed all of the grass in the area covered by #1 and continues on around the building.  What is the new radius when the rope is interrupted by the building? What is that area covered using this new radius of rope before the rope is interrupted by the building again?
A: Ok, I have no idea. This might as well be Greek because that's as much sense as it makes to me. Can someone please explain how this works to me? What do they mean by the radius when the rope is interrupted? I can't wrap my dumb head around this at all. 'Course, I have no idea about the rest of the questions either.

3. What if the horse had gone around the building the other way.  What would the new radius have been when the rope was interrupted by the building?  What is that area covered using this new radius of rope before the rope is interrupted by the building again?

4. The areas you found in 7 and 8 overlap each other. How much do they overlap? What *approximate* shape do they make? What is that area?

5. What is the total area the horse can graze using your calculations from #1-4?

Thanks a bunch!

#6 Re: Help Me ! » Areas, circumferences, sectors of circles » 2021-07-27 20:09:26

Thanks! Sorry, I was being a major tard and missing out on stuff. It's like... 7.07?

#7 Re: Help Me ! » Areas, circumferences, sectors of circles » 2021-07-27 02:26:54

darn, I'm definitely missing something, but I'm too dumb to know what

#8 Re: Help Me ! » Areas, circumferences, sectors of circles » 2021-07-27 02:24:46

Oh wait, so is the angle of the sector 22.5? Since the entire angle is 360, and there are 16 pieces, so 360/16 = 22.5, and now I've got to figure out the radius. But I'm a bit stumped. Gotta have the arc length to find the diameter, but can't find the arc length without the diameter.

#9 Re: Help Me ! » Areas, circumferences, sectors of circles » 2021-07-26 20:57:34

Hi, thanks for the answer. The trick question does me in again!! 314.16 it is.

Still not sure about Q5, though. I'm no good with this stuff at all. I dunno whether what they mean when they say the radius of a half piece. Across the side or up and down?? I've got no clue at all.

#10 Help Me ! » Areas, circumferences, sectors of circles » 2021-07-26 14:33:50

simonmagusflies
Replies: 7

Please help me out with this question.

My mom made a round cake with a 10 inch radius. She cut the cake into 16 equal slices. When 1 slice was left, my sister and I both wanted it, so we agreed to cut it in half, but I like the icing more than she does, so we decided to cut it the "other way." In other words, the two pieces would not be symmetrical. The inside piece would contain less icing, and the outer piece would contain the icing.

https://i.imgur.com/yQ2zLKX.png

1. Find the area of the whole cake.
A: π x 102^2. The area is 314.16

2. What is the area of one piece of cake?
A: I'm not sure what's being asked. Since the cake is split into 16 pieces, you just divide 314.16 by 16, right? That gives 19.635.

3. What is the area of a half-piece?
A: Half of a piece is 19.635 divided by 2? I'm not sure. That gives 9.8175.

4. What would the area of the whole cake be if it were cut into half pieces?
A: Not sure what's being asked.

5. What is the radius of a half-piece when it is cut as shown in the picture? (ie, where do I need to cut to make two equal halves out of a piece?)
A: ∑(´△`○)I have no idea.

Thank you!

#12 Help Me ! » Please help me with trigonometry » 2021-06-27 15:12:22

simonmagusflies
Replies: 3

https://i.imgur.com/aLAIl5k.png

You have climbed to the top of a tall tree. When you get to the top, you use your clinometer to discover that the angle between the tree and the line of sight to your red lunchbox is 30°. You know you left the lunchbox 20 meters from the base of the tree. How tall is the tree?

A. 75.36 m
B. 92.09 m
C. 20.17 m
D. 51.25 m
E. 18.95 m
F. 34.64 m

The last question!! Please help, I'd love to get this done as quickly as I can. Thank you!!

#13 Help Me ! » Missing triangle angles » 2021-06-21 14:08:13

simonmagusflies
Replies: 3

A right triangle has a 90 degree angle, and an angle x. What will the measure of the third angle be?

I answered 45 degrees, but the teacher said: "The angles are not 45 degrees because they are not necessarily equal to each other." There are a bunch of things it could be, then, but what's the "definitive" answer? Thanks.

#14 Re: Help Me ! » Can't wrap my dumb head around this angles problem » 2021-06-08 15:30:06

Nevermind, I think I've got it. P = 70?
You can do p + (90 - p) + (180 - p) = 200, remove the brackets, combine the "p"s, multiply negatives in, which leaves me with p = -200 + 90 + 18. So, that means p = 70?

#15 Re: Help Me ! » Can't wrap my dumb head around this angles problem » 2021-06-08 14:50:01

Bob wrote:

Hi, Your three equations are correct, so this ceases to be an angles question and becomes an algebraic one.
Add together the first two (2p + g + h) and subtract the third to get p.

Bob

Hello, thanks for replying. I just can't wrap my head around that. What am I supposed to do?? (@_@)

#16 Help Me ! » Can't wrap my dumb head around this angles problem » 2021-06-08 00:58:10

simonmagusflies
Replies: 5

An angle "p" exists that is complementary to angle g on one side. On the other side, the angle p is supplementary to angle h. The sum of the three angles is 200°. What is the measure of this common angle p? (Show your work AND upload a drawing of this problem.)

So, p + g = 90
p + h = 180
p + g + h = 200?
Where in the world do I start with this? And the drawing? Aggghhh. I'm really not good with angles.

#17 Re: Help Me ! » Help with parallel angles please » 2021-06-06 22:11:21

Thanks, it was correct! You're terrific!

#18 Re: Help Me ! » Help with parallel angles please » 2021-06-06 14:24:59

So, to both, the answer is "Yes because angles 5/4 and 3/6 are same-side interior angles."?

#19 Re: Help Me ! » Help with parallel angles please » 2021-06-06 01:10:03

Thanks for answering, here are the definitions:

1. Interior angles:
"No matter what the measurements are, angles 3, 4, 5 and 6 are called interior angles because they are in between lines AB and CD."

2. Exterior angles:
"Angles 1, 2, 7 and 8 are called exterior angles. Angles 1 and 8 are called alternate-exterior angles, as are angles 2 and 7."

3. Alternate-interior angles:
"Angles 3 and 6 are called alternate-interior angles, as are angles 5 and 4."

4. Corresponding angles:
"Angles 2 and 6 are called corresponding angles, as are angles 1 and 5, angels 3 and 7, and angels 4 and 8."

5. Vertical angles:
"Also regardless of the measurements, angles 1 and 4 will be the same size, as will angles 2 and 3, angles 5 and 8, and angles 6 and 7.  These are called vertical angles."

When AB is parallel to CD, the following statements are true:
- all corresponding angles are congruent, the same size. (And the corresponding angles are ONLY the same size if the lines are parallel.) Thus by knowing the measurement of 1 angle in the group, you could name the measurements of all the angles in the picture.
- alternate-interior angles are congruent
- corresponding angle are congruent
- alternate-exterior angles are congruent
- same-side interior angles will be supplementary. Angles 3 and 5 in the image above are same-side interior angles. Since angles 3 and 4 are supplementary and angles 4 and 5 are the same measure (alternate-interior angles are congruent), angles 3 and 5 are supplementary. Thus same-side interior angles are supplementary.

#20 Help Me ! » Help with parallel angles please » 2021-06-05 22:52:00

simonmagusflies
Replies: 6

I need help with 2 questions, please.
Here's the angle:

yZlpPMf.jpg

The questions are:
In the figure linked, is AB is parallel to CD...
1. If angle 5 = 30° and angle 3 = 150°?
A. Yes because angles 5 and 3 are same-side interior angles.
B. Yes because angles 5 and 3 are alternate interior angles
C. Yes because angles 5 and 3 are vertical angles.
D. No because angles 5 and 3 are vertical angles.
E. No because angles 5 and 3 are corresponding angles.
F. No because angles 5 and 3 are alternate interior angles

2. If angle 4 = 80° and angle 6 = 100°?
A. Yes because angles 4 and 6 are vertical angles.
B. Yes because angles 4 and 6 are same-side interior angles.
C. Yes because angles 4 and 6 are corresponding angles.
D. No because angles 4 and 6 are vertical angles.
E. No because angles 4 and 6 are alternate interior angles.
F. No because angles 4 and 6 are alternate exterior angles.

I'm pretty sure both answers are one of the "no"s, but I'm not sure of the reason.

#21 Re: Help Me ! » Venn diagram help » 2021-06-03 19:40:14

4 in white intersect long tails and 2 in long tails? It adds up!! Thanks!

#22 Re: Help Me ! » Venn diagram help » 2021-06-03 19:29:51

Sorry if I'm being obtuse, but I can't really figure this out. If I change the 3 in long tails, long tails won't add up to 8 anymore? Do I up one of the threes? My brain is scrambled eggs right now, haha.

#23 Re: Help Me ! » Venn diagram help » 2021-06-03 19:19:47

OK, I've fixed it, is this right?
https://i.imgur.com/Jd3lfS7.png

#24 Help Me ! » Venn diagram help » 2021-06-03 16:38:37

simonmagusflies
Replies: 7

Problem: A cat has thirteen kittens. Eight of the kittens have white hair, six of the kittens have spots, and eight of the kittens have long tails. All of the kittens have at least one of these traits. One kitten is white with spots and a long tail. Three of the kittens are white with spots. Two kittens have spots and long tails. One kitten has white hair but does not have spots or a long tail.

a. Draw a Venn diagram for this problem.
b. How many kittens are white with long tails, but don't have spots?
My answer is 3.

I'm having a hard time wrapping my dull head around this one. So far, I have:
https://i.imgur.com/HrIVYmE.png

But it adds up to 12, not 13. I definitely have something wrong with the spots cats, it can't surely be 0 (maybe?), but it says 6 cats with spots and 3 + 1 + 2 = 6 which leaves no room for cats with just spots.

Please help me out, thanks.

#25 Re: Help Me ! » I'm really at my wit's end with these geometry questions » 2021-05-29 14:32:18

I decided to give it one go and wrote "I have two rays with the same endpoint, which I'll call A. For the first ray, there's a set of points that uses A as an endpoint. Any other point I'll call B.

For the second ray it's about the same, but I'll call any other points C.

So we have A, B, and C, which are three non-collinear distinct points, so we have a plane. That's a point not on the line as well (I guess), and the lines intersect at A."

The teacher replied: "You're on the right track with this thought! The definition of an angle does not include any mention of lines, so you will need to explain where any lines you need come from using what you learned in Lesson 1. Make sure to write a separate proof for each problem as well."

Ahh I wish she'd just mark me off so I can move on!!!

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