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please help cant figure out

\(\begin{array}{|lrcll|} \hline (1) & 3y &=& -x-3 \quad & | \quad :3 \\ &\mathbf{ y} &\mathbf{=}& \mathbf{-\frac13 x-1} \\\\ (2) & 2y-14 &=& 4x \\ & 2y &=& 4x - 14 \\ & \mathbf{y} &\mathbf{=}& \mathbf{2x + 7} \\\\ (3) & 4x-3-y &=& 0 \\ & \mathbf{y} &\mathbf{=}& \mathbf{4x-3} \\\\ (4) & x-12 &=& -3y \\ & 3y &=& -x+12 \quad & | \quad :3 \\ &\mathbf{ y} &\mathbf{=}& \mathbf{-\frac13 x+4} \\ \hline \end{array}\)

A The lines form a aquadrilateral

B The lines (1) and (4) have the same slope.

Since no one answered my last question I'm going to post it again. 

I have 3 pieces of candy to place in 4 lunch boxes.

In how many ways can I do this if exactly two of the candies are the same (but the third is different)

and all of the lunch boxes are different?

Same means indistinguishable and different means distinguishable.

Thanks!

Let candi 1 = sort a

Let candi 2 = sort a

Let candi 3 = sort b

\(\begin{array}{|c|c|c|c|c|} \hline & \text{Box $1$} & \text{Box $2$} & \text{Box $3$} & \text{Box $4$} \\ \hline 1& b & a & a & \\ 2& b & a & & a \\ 3& b & & a & a \\ 4& a & b & a & \\ 5& a & b & & a \\ 6& & b & a & a \\ 7& a & a & b & \\ 8& a & & b & a \\ 9& & a & b & a \\ 10& a & a & & b \\ 11& a & & a & b \\ 12& & a & a & b \\ \hline \end{array} \)

12 ways

Thank you!

Ended up getting there in the end, but wanted to keep the question up so I could get my answer confirmed.

Thanks for your help.

First, you start with writing out the triangle and the givens.

Angle Q = Angle Q -> Angles equal themselves

QM congruent QP -> QM is half of QP since M is the midpoint, so they're congruent

QN congruent QR -> Same reason as one above

Triangle QMN congruent Triangle QPR -> SAS

MN || PR -> Same Side Rule (Forgot what it's called exactly)

There ya have it

The frequency is B in the equation \(y=Asin(B(x-C))+D\) . In your equation, there's no C, so \(\frac{x}{4}=B\).

Just for anyone who doesn't want to draw the pic. ;)

First, all equations need to equal y.

\(3y=-x-3\)

\(y=\frac{x-3}{3}\)

----

\(2y-14=4x\)

\(2y=4x+14\)

\(y=2x+7\)

----

\(4x-3-y=0\)

\(y=4x-3\)

----

\(-3y=x-12\)

\(y=\frac{x-12}{-3}\)

Next, plug all equations into a graphing calculator. It should look something like this. (I've made the lines thicker and made the points orange)

Aaaaaand I don't know what b means. Eh, 1 out of 2 isn't that bad ¯\_(ツ)_/¯

As EP said above, it's hard to do these types of problems without an image to see. So, let's make an image.

He starts out with the dice, so he has a chance of 1-6.

When he rolls a 1 or a 2, he goes to the coin. Otherwise, he stays on the dice.

Now we can write which corresponds to what Bill's next turn will result in coin or dice

Now that there's an image, we can count how many chances go to the coin on that last turn.

There are 14 total outcomes by the 3rd move. 5 of them lead to using the coin on the last turn.

The probability of Billy using a coin on the last turn should be 5/14.

I don't know if this is right, since EP had 2 totally different answers than I got, so... xD

Use exponential growth for this one as well:

15 x [2^N - 1] =

15 x [2^9 - 1] =

15 x [512 - 1] =

15 x 511 =

=7,665

You can also use GS to get the same answer =7,665

What you're going to want to do is something like \((15*2^0)+(15*2^1)+(15*2^2)+(15*2^3)+(15*2^4)+(15*2^5)\)\(+(15*2^6)+(15*2^7)+(15*2^8) = 7665​\). I know it's long. Hold on.

You're basically going to do \(1+2+4+8+...\), but you're going to multiply every number by 15 since they sold 15 arrangements in the first month.

I hope this explanation makes sense. ¯\_(ツ)_/¯

In this case, you would do \(2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11+ = 4095\). Since it doubles, it makes sense to use 2 as a base. Since it starts with one car, you start with a value that equals 1 (2⁰) and continues all the way to 12 months later (2¹¹ ). You add all these up since you're asking for the total amt of cars spent in the first year.

You can use exponential growth for this one:

2^N - 1, where N = Number of months

2^12 - 1

4,096 - 1

=4,095 - total number of cars sold in 12 months!!. Quite a dealership!!

You can also use GS to sum them up, where:

First term =1

Common ratio = 2

Number of terms = 12

You should get the same answer =4,095 cars.

Well, you need to find the surface area of the cake. If you fold the prism out, you can find each side and add it out. You can fold it out like so.

You should then find the areas of all the sides except the middle square. The reason for this is because the middle square represents the bottom of the cake and you don't ice the bottom.

Then add all the areas of each square.

45+45+150+30+30 = 300 in²

4 + 12 + 36 + 108 + 324 + 972 + 2,916 + 8,748 =13,120

You can also sum it up as GS =13,120

Use the same formula for Geometric Series

First term = -600

Common Ratio= 1/2

Number of terms =8

You should get =- 1,195.31

Thank you so much!

This is a Geometric Series, where:

F = First term = -8

R = Common Ratio = - 2.

N =Number of terms =7. Use the formula for GS

S = F x (1 -R^N) / ( 1 - R)

S = -8 x (1 - (-2^7)) / (1 - (- 2))

S = -8 x (1 - (-128)) / (1 + 2)

S = -8 x (1 + 128) / 3

S = -8 x (129 / 3)

S = -8 x 43

S = - 344

its all beings times by 3.

Soo

4*3=12

12*3= 36

36*3=108

108*3=324

324*3=648

648*3=1296

1296*3=3888

make sense so far.. But ill double check myself.

-Hayley1

About 12.9 minutes

b) 75.8 feet

c) 45 degrees

If you want the answer to be 6, then it is written wrong! It should read like this:

 -6 - [ (-16) / (4/3) ]

-6 - [ (-16) x (3/4) ]

-6 - [ -48/4 ]

-6 - [ - 12]

-6 + 12

= 6

a) tan(theta/2) = x/2d.

b) not sure what the question is asking....but if you need to calculate x then it's approx 0.007 or d is 9167.32 for the given angle

https://web2.0calc.com/questions/what-constant-k-will-make-y-2-24y-16-k-the-square-of-a-binomial

Bam! I remember seeing this. 

-Hayley1

It's suppose to be 6. Cannot figure out why.

-Hayley1

-6-(-16) Divded by 4/3

[- 6 + 16] / (4/3)

[+10] x (3/4)

30 / 4

=7 1/2