we have been asked to simplify
![(3x-5)-(5x + 1)](https://tex.z-dn.net/?f=%20%20%283x-5%29-%285x%20%2B%201%29%20)
To Simplify this, we will first open the parenthesis, as below
![(3x-5)-(5x + 1)=3x-5-5x-1\\ \\ \text{take the like terms together and simplify we get}\\ \\ (3x-5)-(5x + 1)=3x-5x-5-1\\ \\ (3x-5)-(5x + 1)=-2x-6\\ \\\\](https://tex.z-dn.net/?f=%20%283x-5%29-%285x%20%2B%201%29%3D3x-5-5x-1%5C%5C%20%5C%5C%20%5Ctext%7Btake%20the%20like%20terms%20together%20and%20simplify%20we%20get%7D%5C%5C%20%5C%5C%20%283x-5%29-%285x%20%2B%201%29%3D3x-5x-5-1%5C%5C%20%5C%5C%20%283x-5%29-%285x%20%2B%201%29%3D-2x-6%5C%5C%20%5C%5C%5C%5C%20)
Hence the simplified expression is -2x-6
Answer:
smaller number = 15
Step-by-step explanation:
Let the two numbers be 5x and 3x respectively.
According to the question —
5x — 3x = 10
2x = 10
x = 5
Thus the smaller number is 3 × 5 = 15
Answer:
7. C. 6
8. H. √34
9. A. (1, 3.5)
10. J. 10
Step-by-step explanation:
7. AB = 2y, BC = 6y, AC = 48
AB + BC = AC (segment addition theorem)
Substitute the above values into the equation
2y + 6y = 48
Solve for y
8y = 48
Divide both sides by 8
8y/8 = 48/8
y = 6
8. Distance between P(2, 8) and Q(5, 3):
![PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}](https://tex.z-dn.net/?f=%20PQ%20%3D%20%5Csqrt%7B%28x_2%20-%20x_1%29%5E2%20%2B%20%28y_2%20-%20y_1%29%5E2%7D%20)
Let,
![P(2, 8) = (x_1, y_1)](https://tex.z-dn.net/?f=%20P%282%2C%208%29%20%3D%20%28x_1%2C%20y_1%29%20)
![Q(5, 3) = (x_2, y_2)](https://tex.z-dn.net/?f=%20Q%285%2C%203%29%20%3D%20%28x_2%2C%20y_2%29%20)
![PQ = \sqrt{(5 - 2)^2 + (3 - 8)^2}](https://tex.z-dn.net/?f=%20PQ%20%3D%20%5Csqrt%7B%285%20-%202%29%5E2%20%2B%20%283%20-%208%29%5E2%7D%20)
![PQ = \sqrt{(3)^2 + (-5)^2}](https://tex.z-dn.net/?f=%20PQ%20%3D%20%5Csqrt%7B%283%29%5E2%20%2B%20%28-5%29%5E2%7D%20)
![PQ = \sqrt{9 + 25}](https://tex.z-dn.net/?f=%20PQ%20%3D%20%5Csqrt%7B9%20%2B%2025%7D%20)
![PQ = \sqrt{34}](https://tex.z-dn.net/?f=%20PQ%20%3D%20%5Csqrt%7B34%7D%20)
9. Midpoint (M) of segment LB, for L(8, 5) and B(-6, 2) is given as:
![M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})](https://tex.z-dn.net/?f=%20M%28%5Cfrac%7Bx_1%20%2B%20x_2%7D%7B2%7D%2C%20%5Cfrac%7By_1%20%2B%20y_2%7D%7B2%7D%29%20)
Let ![L(8, 5) = (x_1, y_1)](https://tex.z-dn.net/?f=%20L%288%2C%205%29%20%3D%20%28x_1%2C%20y_1%29%20)
![B(-6, 2) = (x_2, y_2)](https://tex.z-dn.net/?f=%20B%28-6%2C%202%29%20%3D%20%28x_2%2C%20y_2%29%20)
Thus:
![M(\frac{8 + (-6)}{2}, \frac{5 + 2}{2})](https://tex.z-dn.net/?f=%20M%28%5Cfrac%7B8%20%2B%20%28-6%29%7D%7B2%7D%2C%20%5Cfrac%7B5%20%2B%202%7D%7B2%7D%29%20)
![M(\frac{2}{2}, \frac{7}{2})](https://tex.z-dn.net/?f=%20M%28%5Cfrac%7B2%7D%7B2%7D%2C%20%5Cfrac%7B7%7D%7B2%7D%29%20)
![M(1, 3.5)](https://tex.z-dn.net/?f=%20M%281%2C%203.5%29%20)
10. M = -10, N = -20
Distance between M and N, MN = |-20 - (-10)|
= |-20 + 10| = |-10|
MN = 10
Major arc, because it's bigger than a semicircle