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Hey guest! In this problem, we're going to get involved with some systems of equations. First, we have that Allysa bought two different types of chickens: Americana and Delaware, Let's name the quantity that she purchased of each of these chickens \(a \) and \(d\) respectively. We have that she purchased 12 chickens total, so we can write the equation:

\(a + d = 12\)

The problem then tells us that Americana chickens cost $3.75 each, and Delaware Chickens cost $2.50 each. It also gives us that the total amount of money that Allysa spent was 35$. We can then write:

\(3.75a + 2.50d = 35\)

The reasoning for this equation is because each Americana is 3.75, so 3.75 * the number of Americanas purchased represents the total amount of money she spent on Americanas, and likewise for Delaware chickens. 

Now, let's multiply our first equation by 2.50, and then subtract it from our second equation to get the value of a.

We get:

\(2.50a + 2.50d = 12 * 2.5 = 30\)

Subtract this from our second equation, and we get:

\(1.25a = 5\)

Dividing by 1.25 on both sides, we get:

\(a = \frac5{1.25} = 4\)

Now that we have the number of Americans purchased, a, we can just plug this back into our first equation \(a+d = 12\)

\(4 + d = 12\)

Subtracting 4 on both sides, we get:

\(d = 8 \)

Hey guest! I'm assuming you meant \(37 \frac14 * 32\) for this problem.

To get the answer for this one, I'd imagine there are a variety of solutions, but rather than converting \(37 \frac14\) into a fraction and then multiplying that by 32(that would be quite cumbersome in my opinion), what I would do is split the mixed number into two parts. A whole number, and a fractional part(The reasoning behind this is not random!).

We can write the following equivalence:

\(37 \frac14 * 32 = (37 + \frac14) * 32\)

We then distribute this into:

\((37 + \frac14) * 32 = (37*32) + (\frac14 * 32)\)

Do you see now why this method makes this particular problem easy? If you haven't noticed, the right side, which is \((\frac14 * 32)\), reduces very nicely, as 4 is a factor of 32(32 is 4 * 8). This leaves us with:

\((37 * 32) + (8) \)

Plugging this in a calculator, this converts to:

\((1184) + (8) = 1192\) as our final answer

Hey guest! Let's start with some basic background knowledge of types of interest:

there are two main types, simple and compound. With simple interest, given a quantity:

\(x\), and in interest rate of \(r\%\) per year, we have that every year, the interest amount is:

^{\(x * r\%\)}. In other words, the amount paid every year(if we're talking about an interest on payment) is the exact same constant value.

Compound interest however, given the same variables, is a bit different.

Compound interest is "compounded"(hence the name) and added on to the previous year(or whatever metric of time used). The formula for compound interest is then typically defined:

\(A = P(1+\frac{r}{n})^{nt}\)

Where A = Amount paid

P = The principal, or the initial amount first paid 

R = Interest rate as a decimal

N = The number of times the interest is compounded per unit of time "t"

T = Time(in this example, years).

Here's a great website for reference on compound interest:

https://www.thecalculatorsite.com/articles/finance/compound-interest-formula.php

Getting to the problem itself:

A)

The interest that Sue gets with an investment of 2300 pounds and an interest rate of 2.4% a year over 3 years is:

\(A = 2300(1+\frac{0.024}1)^{3*1} = 2300(1.024)^3 = 2300(1.073741824) = 2469.6061952\)

For Bill, he invests 1800 pounds at an interest of 3.4% per year.

Our equation is then:

\(A = 1800(1+\frac{0.034}1)^{3*1} = 1800(1.034)^3 = 1800(1.105507304) = 1989.9131472\)

Of course, both of these answers are using calculators.

B)

To solve this, we figure out the compound interest for Bill after 2 years, and then multiply that amount by (1 + 0.04), which represents the 4 percent compound interest rate added on to that.

Sue:

We have already that Sue has an interest payment of 2469.6061952 after 3 years

Now, moving on to Bill:

By our previous formula:

\(\)

\(A = 1800(1.034)^2 = 1800(1.069156) = 1924.4808\)

We then multiply this value by 1.04 to represent a 4 percent interest rate(4% = 0.04, so 1+ 0.04 represents a 4 percent compounded interest).

\(1924.4808 * 1.04 = 2001.460032\)

Clearly, this is less than Sue's amount after 3 years, so the answer is then Sue

Example...I'll do the cement   he has 8 bags     8 bags x 25 kg/bag = 200 kg     He needs 180 kg....he does not need to purchase any more cement.

                                 sand                   19 bags   19 bags X 22.5 kg/bag =......     etc etc

2 w   +  12c    =  33

5 w   +   3 c    =15        Multiply this by -4

-20 w  - 12 c = -60        Add to top equation

-18w             = - 27

w = $1.50/lb     I think you can now calc the price of  w   .....

-GA

First of all, if I was really talking from an unknowledgeable perspective, what is your rationale behind me not noticing the logical fallacy presented by Cal? This makes no sense whatsoever. Cal tells us that :

"

The largest power of 2 that divides 20! is 2^18 .

The largest power of 3 that divides 20! is 3^8. 

We want the largest value of n that n is less than or equal to 8.

"

If I really had no knowledge of Legendre's formula like you so eagerly recommended, how would I have known anything about this kind of reasoning? If I was really uneducated and was looking at the steps Cal used as opposed to just the answer, I would've said "wait, how does this logic follow through"? Because if I didn't know anything about the subject, then I wouldn't have understood those last two lines at all. Your assumption here is just blatantly false.

Secondly, sure, say I buy your argument that there are many solutions on this forum presented from a knowledgeable perspective. Does that mean we should present all of our solutions from "a perspective that requires knowing the answer"? No, it does not by any means make us assume that our question-askers know what we're talking about. Rather, we should go through each step and explain the logic behind it. I'd argue that this is comparatively better than making them "learn the rote mechanics, and then the logic behind the process" themselves as you advocate for, because when we show question-askers the logic and decision-making calculus behind our steps, that allows them to go forward with more knowledge to tackle different types of these problems, which they can always combine with a bit of self-learning. We shouldn't just post the answer with logical leaps and expect our readers to understand. 

Thirdly, it was a mistake on my part to not fully explain Legendre's as you mentioned, however, I'm not trying to rip "Cal a new one for her incompetence", I'm trying to help her so she doesn't post misleading solutions that she doesn't fully understand; that just causes confusion for the question askers and everyone in general. 

That is a loooong way away Tacobell.

Maybe you should have a 12.5 celebration.  You are probably in lock down so Maybe it will have to be on the net.

HAPPY HALF BIRTHDAY     (It is just a pity your birthday, or half birthday is not in Juni .. That is German for June.)

Don't eat it all at once   

Total students = 870 +800 = 1670

2/3  * 870 = boys spanish = 580

2/5  * 800 = girls                 = 320     total 900 who study spanish out of  1670 total students

                                                                   1670 - 900 = 770 out of 1670 who do not study spanish    770/1670 = 77/167

he meant 100

Well, correrct me if im wrong, 

a pound of walnut = $1.5

a pound of choco chips = $2.5

Calculate the perimeter of the shape.

Deleted, there was an error

Work out the length of AB .

To the nearest degree, determine all possible values of θ in the domain -360° ≤ θ ≤ 360° given cos θ = 1/3.

As follows:

Ok Cat,  I do not think you got your answer from that site.

But what you have written could not be understood by you. 

I know that because it cannot be understood by me.

If you do not understand an answer, you should not print it

OR you should admit straight away that you had help from someone else, go on to say that you believe it to be true, but that you cannot explain it yourself because you do not really understand.

In trapezoid \(PQRS\), \(\overline{PQ} \parallel \overline{RS}\).
Let \(X\) be the intersection of diagonals \(\overline{PR}\) and \(\overline{QS}\).
The area of triangle \(PQX\) is \(20\), and the area of triangle \(RSX\) is \(45\).
Find the area of trapezoid \(PQRS\).

\(\begin{array}{|rcll|} \hline \text{Let Area }A_1 &=& [PQX] \\ A_1 &=& 20 \\\\ \text{Let Area }A_2 &=& [RSX] \\ A_2 &=& 45 \\\\ \text{Let Area }A_3 &=& [PXS] \\\\ \text{Let Area }A_4 &=& [QXR] \\\\ \text{Let Area }A_5 &=& [PSR] \\ &=& A_2+A_3 \\\\ \text{Let Area }A_6 &=& [QSR] \\ A_6 &=& A_2 +A_4 \\ \hline \mathbf{A_5} &=& \mathbf{A_6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A_5} &=& \mathbf{A_6} \\ A_2+A_3 &=& A_2 +A_4 \\ \mathbf{A_3 }&=& \mathbf{A_4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A_3 = \dfrac{SX*h_1}{2} && A_1 = \dfrac{QX*h_1}{2} \\ \mathbf{\dfrac{A_3}{A_1} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ A_2 = \dfrac{SX*h_2}{2} && A_4 = \dfrac{QX*h_2}{2} \\ \mathbf{\dfrac{A_2}{A_4} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ \dfrac{SX}{QX}=\mathbf{\dfrac{A_3}{A_1} }&=&\mathbf{\dfrac{A_2}{A_4} } \quad | \quad A_4 = A_3 \\\\ \dfrac{A_3}{A_1} &=& \dfrac{A_2}{A_3} \\\\ A_3^2 &=& A_1A_2 \\ A_3^2 &=& 20*45 \\ A_3^2 &=& 900 \\ \mathbf{A_3} &=& \mathbf{30} \\\\ A_4 &=& A_3 \\ \mathbf{A_4} &=& \mathbf{30} \\ \hline \end{array}\)

\(\text{The area of trapezoid $PQRS = A_1+A_2+A_3+A_4 = 20+45+30+30=\mathbf{125}$ }\)

Your cut a penny joke is funny, SpongeBob.

You are also correct that the result should be calculated to the nearest cent. All the data is presented to the nearest cent; using more significant digits is superfluous.  If the question indicated this was to be used as a statistic for a chain of a thousand plus restaurants, then the third decimal place would be significant in that collective.   

Jfans analogy comparing the use of significant digits to that of a hypothetical question where the construct could not or is not likely to exist in the real world is non sequitur.  Hypothetical questions are common in mathematics and physics. For example, physics questions will often say to ignore air resistance or other frictions that would normally be present in reality. The purpose of this is to quantize the learning process. There are also real environments where such frictions are insignificant and all of these question types have an at least an implied final significant digit count.    There usually is not a need to carry significant digit use to extremes. So, how many digits of /pi do we need?

--------------------

Hypothetical future conversation... sometime after SpongeBob reattaches his corpus spongiosum appendage handed to him by jfran.

Jfan17: The problem is you are a generalist and I am a perfectionist.

SBR24: Whatever...

Jfan17: Exactly!

GA

Let a and b be positive real numbers such that a + b = 1. Find set of all possible values of 1/a + 1/b.

Hello Guest!

\(a+b=1\\ a=1-b\\ \frac{1}{a}+\frac{1}{b}= \frac{1}{1-b}+\frac{1}{b}=\frac{b+1-b}{b-b^2}=\frac{1}{b-b^2}\\ \)

The set of all possible values of   \(\frac{1}{a}+\frac{1}{b}\)  are \(\{\mathbb R>1\}\ |\ \{a,b\} \subset \mathbb R\ \)und 1> a,b > 0

  !

nchang,  behave yourself.

You post an incomplete question and then deduct points from the person who tells you that.

You should be pleased that people show interest in your question

And you should apologise for wasting people's time by not proof reading it in the first place.

There is another person here who should also behave themselves.

Yes, I do know who you are!     

jfan is right. Some of the problem is missing.

But here is the pic that you need.

Maybe not  the most elegant way to do this....but  .....

20!  =

20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2

(2 * 2 * 5) (19) ( 3 * 3 * 2) * 17 * (2 * 2 * 2 * 2) * (3 * 5)  * ( 7 * 2) * (13) * (12) * (11) * (2 *5)  * (3 * 3) * (2 * 2 * 2) * (7) *

(2 * 3)  * ( 5) * ( 2 * 2) * ( 3 * 2)

In addition to 12, we only need consider  powers  of 2  and  3  

[ ( 2 * 2 )  ( 3 * 3 * 2 ) ( 2 * 2 * 2 * 2)  (3 * 2 * 12 * 2 ) (3 * 3 * 2 * 2*  2) ( 2 * 3 * 2 * 2 * 3 * 2) ]  =

[ 2 ^16  *  3^7  * 12 ]     { Break  this up into  powers of 12 }    

[ 3 * 2^2 ]  [ 3 * 2^2 ] [ 3 * 2^2 ]  [ 3 * 2^2 ] [3 *2^2 ] [3 * 2^2]  [ 3 * 2^2 ]  * 12   *   [ 2^2] ignore this

     12  *      12     *     12     *     12     *    12   *    12    *    12     *   12

12^8   will  divide 20!

So

n =  8