Answer:
she should add a bus , because u need to have enough buses for all students, u cant just leave some behind
Step-by-step explanation:
Answer:
The area of the smallest section is ![A_{1}=100yd^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%3D100yd%5E%7B2%7D)
The area of the largest section is ![A_{2}=625yd^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%3D625yd%5E%7B2%7D)
The area of the remaining section is ![A_{3}=250yd^{2}](https://tex.z-dn.net/?f=A_%7B3%7D%3D250yd%5E%7B2%7D)
Step-by-step explanation:
Please see the picture below.
1. First we are going to name the side of the larger square as x.
As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:
- Area of the first section:
![A_{1}=10yd*10yd](https://tex.z-dn.net/?f=A_%7B1%7D%3D10yd%2A10yd)
![A_{1}=100yd^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%3D100yd%5E%7B2%7D)
- Area of the second section:
(Eq.1)
- Area of the third section:
![A_{3}=width*length](https://tex.z-dn.net/?f=A_%7B3%7D%3Dwidth%2Alength)
(Eq.2)
2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:
![A_{1}+A_{2}+A_{3}=975yd^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%2BA_%7B2%7D%2BA_%7B3%7D%3D975yd%5E%7B2%7D)
Replacing values:
![100+x^{2}+10x=975](https://tex.z-dn.net/?f=100%2Bx%5E%7B2%7D%2B10x%3D975)
Solving for x:
![x^{2}+10x-875=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B10x-875%3D0)
![x=\frac{-10+\sqrt{100+(4*875)}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-10%2B%5Csqrt%7B100%2B%284%2A875%29%7D%7D%7B2%7D)
![x=\frac{-10+\sqrt{3600}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-10%2B%5Csqrt%7B3600%7D%7D%7B2%7D)
![x=\frac{-10+60}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-10%2B60%7D%7B2%7D)
![x=25](https://tex.z-dn.net/?f=x%3D25)
3. Replacing the value of x in Eq.1 and Eq.2:
- From Eq.1:
![A_{2}=25^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%3D25%5E%7B2%7D)
![A_{2}=625yd^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%3D625yd%5E%7B2%7D)
- From Eq.2:
![A_{3}=10*25](https://tex.z-dn.net/?f=A_%7B3%7D%3D10%2A25)
![A_{3}=250yd^{2}](https://tex.z-dn.net/?f=A_%7B3%7D%3D250yd%5E%7B2%7D)
So to solve for this, we need to set up proportional fractions, which I will help show you how to do.
First, if we are given an amount out of a total, we need to put it over x (if we are looking for the total). It looks like this:
12/x, 12 being the given number and x being the total.
If we are given the total but are looking for an amount, put the total at the bottom of the fraction (aka the denominator). It looks like this: x/16, 16 being the total amount and x being the amount out of the total.
We have a total of 40 test problems, so we can put our total at the bottom, x/40.
X is the amount of questions answered correctly (we are looking for x in the question).
We have answered 80% correct, so put 80% over 100 (100 being the total). It should look like this: 80/100.
Now we have our two fractions: x/40 & 80/100.
Set these up as an equation.
x/40 = 80/100.
Now this is where things may get tricky if you don't pay attention.
Multiply the numerator (the top number of a fraction) of x/40 by the denominator (the bottom number of a fraction) of 80/100.
Your product equation should look like this:
x times 100. This will give is 100x. Leave it at that.
Now, multiply the denominator of x/40 (the bottom number of the fraction) by the numerator (the top number of a fraction) of 80/100. It should look like this:
80 x 40. This will give us 3200.
Now set up our products as an equation.
100x = 3200.
To solve for x, divide both sides by 100.
3200/100 = 32.
x = 32.
I hope this helps and has taught you something!
Since it includes 11% sales tax, this means that (purchase)+11% of (purchase)=25.00. Since 11% = 0.11 by multiplying it by 1/100, we have 1*purchase+0.11*purchase=1.11*purchase=25.00. Dividing both sides by 1.11, we get around 22.73 as the original purchase. This means that the difference between the purchase price and the total price is the sales tax, which is around 2.27 dollars.
Feel free to ask further questions!