Answer:
23. a-T b-F
24. Words- *3* times the number of *feet*
Variable- Let *y* represent the number of *yards*
Model- (I can't do it)
Expression- The number of feet in*3* yards is given by the expression *3y*
25. 7.8
26. 8.3
27. 14.5
28. Yes. The communitive property tells us that we can switch any two numbers in a multiplication or addition problem
*I did that page like a month ago*
Answer:
D. 3/2
Step-by-step explanation:
The scale factor is 1/2 from OP / NQ = 1/2. Because of that, MQ is equal to half of MP and QP is equal to half of MP. Therefore, QP = MQ = 2. The slope of MO is rise over run which is 6 / 4 or 3 / 2.
Answer:
See Explanation
Step-by-step explanation:
Given
![Total = x^2 + 14x + 24](https://tex.z-dn.net/?f=Total%20%3D%20x%5E2%20%2B%2014x%20%2B%2024)
![Area\ I = x^2](https://tex.z-dn.net/?f=Area%5C%20I%20%3D%20x%5E2)
![Area\ IV = 24](https://tex.z-dn.net/?f=Area%5C%20IV%20%3D%2024)
Required
True about area II and III
The question is incomplete; so, I will answer using general knowledge.
If the total area is:
![Total = x^2 + 14x + 24](https://tex.z-dn.net/?f=Total%20%3D%20x%5E2%20%2B%2014x%20%2B%2024)
And there are 4 areas, then;
![Total = Area\ I + Area\ II + Area\ III + Area\ IV](https://tex.z-dn.net/?f=Total%20%3D%20Area%5C%20I%20%2B%20Area%5C%20II%20%2B%20Area%5C%20III%20%2B%20Area%5C%20IV)
Rewrite as:
![Total = Area\ I + Area\ IV+ Area\ II + Area\ III](https://tex.z-dn.net/?f=Total%20%3D%20Area%5C%20I%20%2B%20Area%5C%20IV%2B%20Area%5C%20II%20%2B%20Area%5C%20III)
This gives:
![x^2 + 14x + 24 = x^2 + 24+ Area\ II + Area\ III](https://tex.z-dn.net/?f=x%5E2%20%2B%2014x%20%2B%2024%20%3D%20x%5E2%20%2B%2024%2B%20Area%5C%20II%20%2B%20Area%5C%20III)
Collect like terms
![x^2 -x^2 + 14x + 24 -24= Area\ II + Area\ III](https://tex.z-dn.net/?f=x%5E2%20-x%5E2%20%2B%2014x%20%2B%2024%20-24%3D%20%20Area%5C%20II%20%2B%20Area%5C%20III)
![14x = Area\ II + Area\ III](https://tex.z-dn.net/?f=14x%20%3D%20%20Area%5C%20II%20%2B%20Area%5C%20III)
Rewrite as:
![Area\ II + Area\ III = 14x](https://tex.z-dn.net/?f=Area%5C%20II%20%2B%20Area%5C%20III%20%20%3D%2014x)
<em>So, the sum of areas III and IV must be 14x</em>
M(x) = 5x + 4 n(x) = 6x - 9
Part 1 (m + n)(x) = 5x + 4 + 6x - 9
= 11x - 5
Part 2 (m * n)(x) = (5x + 4)(6x - 9) = 30x^2 - 21x - 36
Part 3 m[n(x)] = 5(6x - 9) + 4
= 30x - 45 + 4
= 30x - 41