$7.56 is the answer to your question
Answer:
Moira didn't add
to both sides of the equation in the fourth step she applied (this is:
).
She can correct it applying the Addition property of equality and add
to both sides of the equation then she must apply the Division property of equality (See the procedure below).
Step-by-step explanation:
Given the equation:
![12x + 10 = 54 - 10x](https://tex.z-dn.net/?f=12x%20%2B%2010%20%3D%2054%20-%2010x)
The steps to solve it are:
1. According to the Subtraction property of equality, you we can subtract 10 from both sides of the equation:
![12x + 10-10 = 54 - 10x-10\\\\12x = 44 - 10x](https://tex.z-dn.net/?f=12x%20%2B%2010-10%20%3D%2054%20-%2010x-10%5C%5C%5C%5C12x%20%3D%2044%20-%2010x)
2. Based on the Addition property of equality, you we can add
to both sides of the equation:
![12x +10x= 44 - 10x+10x\\\\22x=44](https://tex.z-dn.net/?f=12x%20%2B10x%3D%2044%20-%2010x%2B10x%5C%5C%5C%5C22x%3D44)
3. Finally, applying the Division property of equality, we must divide both sidesby 22:
![\frac{22x}{2}=\frac{44}{22}\\\\x=2](https://tex.z-dn.net/?f=%5Cfrac%7B22x%7D%7B2%7D%3D%5Cfrac%7B44%7D%7B22%7D%5C%5C%5C%5Cx%3D2)
Therefore, she made an error in fourth step she applied:
Because she didn't add
to both sides of the equation.
She can correct it applying the Addition property of equality and add
to both sides of the equation and then she must apply the Division property of equality (As you can observe in the procedure shown above).
The most bouquets possibke would be 9 because 45÷9=9 and 63÷7=9
Answer:
There are 59,049 different ways of answering the exam.
Step-by-step explanation:
The first thing we need to count is the total number of selections, then we need to count the number of options for each one of these selections. The total number of combinations will be equal to the product between all the numbers of options.
Here, the selections are each one of the 10 questions, so we have 10 selections.
And each one of these selections has 3 options (true, false, blank)
So, we have 10 selections with 3 options each, then the total number of combinations will be:
C = 3*3*3*3*3*3*3*3*3*3 = 3^10
C = 59,049
There are 59,049 different ways of answering the exam.