For a parallelogram, the area is calculated by the equation,
A = bh
where A is area, b is base, and h is height. From this equation, we can solve for the base of the banner by dividing the area by the height.
base of the banner = 127.5 cm² / 4.25 cm
base of the banner = 30 cm
Thus, the measure of the base of the banner is equal to 30 cm.
<h2>
Hello!</h2>
The answer is: The correct dose of medication to give is 5.6 mL
<h2>
Why?</h2>
To solve this problem we need to establish a relationship between the prescripted medication and the available solution.
Let's write the needed equations to establish the relantionship:
![\frac{125mg}{5mL}=1\\\\125mg=5mL](https://tex.z-dn.net/?f=%5Cfrac%7B125mg%7D%7B5mL%7D%3D1%5C%5C%5C%5C125mg%3D5mL)
The available solution means 125 mg each 5 mL of solution, so:
![\frac{125mg}{5mL}=\frac{140mg}{x}\\\\x*\frac{125mg}{5mL}=140mg\\\\x=140mg*\frac{5ml}{125mg}=\frac{700mg*mL}{125mg}=5.6mL](https://tex.z-dn.net/?f=%5Cfrac%7B125mg%7D%7B5mL%7D%3D%5Cfrac%7B140mg%7D%7Bx%7D%5C%5C%5C%5Cx%2A%5Cfrac%7B125mg%7D%7B5mL%7D%3D140mg%5C%5C%5C%5Cx%3D140mg%2A%5Cfrac%7B5ml%7D%7B125mg%7D%3D%5Cfrac%7B700mg%2AmL%7D%7B125mg%7D%3D5.6mL)
Hence, the correct dose of medication to give is 5.6 mL of the 125 mg/5mL solution.
Have a nice day!
Answer:
Part A:
![4 + x = y](https://tex.z-dn.net/?f=4%20%2B%20x%20%3D%20y)
![0.5*4 + 1*x = 0.7*y](https://tex.z-dn.net/?f=0.5%2A4%20%2B%201%2Ax%20%3D%200.7%2Ay)
![x = 8/3\ quarts\ of\ 100\%\ solution](https://tex.z-dn.net/?f=x%20%3D%208%2F3%5C%20quarts%5C%20of%5C%20100%5C%25%5C%20solution)
![y = 20/3\ quarts\ of\ 70\%\ solution](https://tex.z-dn.net/?f=y%20%3D%2020%2F3%5C%20quarts%5C%20of%5C%2070%5C%25%5C%20solution)
Part B:
![x + y = 10](https://tex.z-dn.net/?f=x%20%2B%20y%20%3D%2010)
![2x + 3y = 23](https://tex.z-dn.net/?f=2x%20%2B%203y%20%3D%2023)
and ![y = 3](https://tex.z-dn.net/?f=y%20%3D%203)
Step-by-step explanation:
Part A:
The inicial concentration of the lemonade is 50%, and the volume is 4 quarts, and we will add x quarts of a lemonade with a concentration of 100%, so the total volume will be y, and the concentration will be 0.7, so we have that:
![4 + x = y](https://tex.z-dn.net/?f=4%20%2B%20x%20%3D%20y)
![0.5*4 + 1*x = 0.7*y](https://tex.z-dn.net/?f=0.5%2A4%20%2B%201%2Ax%20%3D%200.7%2Ay)
Using the value of y from the first equation in the second one, we have:
![2 + x = 0.7*(4 + x)](https://tex.z-dn.net/?f=2%20%2B%20x%20%3D%200.7%2A%284%20%2B%20x%29)
![2 + x = 2.8 + 0.7x](https://tex.z-dn.net/?f=2%20%2B%20x%20%3D%202.8%20%2B%200.7x)
![0.3x = 0.8](https://tex.z-dn.net/?f=0.3x%20%3D%200.8)
![x = 8/3\ quarts](https://tex.z-dn.net/?f=x%20%3D%208%2F3%5C%20quarts)
![y = 4 + 8/3 = 20/3\ quarts](https://tex.z-dn.net/?f=y%20%3D%204%20%2B%208%2F3%20%3D%2020%2F3%5C%20quarts)
Part B:
If he shoots a total of ten targets, we can write the equation:
![x + y = 10](https://tex.z-dn.net/?f=x%20%2B%20y%20%3D%2010)
Each stationary target is 2 points, and each moving target is 3 points, so if the total points is 23, we have:
![2x + 3y = 23](https://tex.z-dn.net/?f=2x%20%2B%203y%20%3D%2023)
If we subtract the second equation by two times the first one, we have:
![2x + 3y - 2*(x + y) = 23 - 2*10](https://tex.z-dn.net/?f=2x%20%2B%203y%20-%202%2A%28x%20%2B%20y%29%20%3D%2023%20-%202%2A10)
![2x + 3y - 2x - 2y = 23 - 20](https://tex.z-dn.net/?f=2x%20%2B%203y%20-%202x%20-%202y%20%3D%2023%20-%2020)
![y = 3](https://tex.z-dn.net/?f=y%20%3D%203)
⇒ ![x = 7](https://tex.z-dn.net/?f=x%20%3D%207)