Answer:
a(n) = a(1) -3(n-1)
Step-by-step explanation:
Given
f(x)= -3x - 5
g(x)= 4x - 2
Find
(f - g)(x)
substitute the f and g values in (f - g)
(f - g)(x)= -3x - 5 - (4x - 2)
change the signs in parentheses because there is a negative sign in front of the parentheses
(f - g)(x)= -3x - 5 - 4x + 2
combine like terms
(f - g)(x)= -7x - 3
ANSWER: (f - g)(x)= -7x - 3
Hope this helps! :)
Answer:
First one is a, second one is d, and the third one is b
Step-by-step explanation:
hope this helps friend
So basically an arithmetic sequence has a common difference, a number which is either added or subtracted at a constant rate (only that number). A geometric sequence is the ratio between two numbers, meaning they are either multiplied or divided by the same number. The sequence would be neither if it follows none of these patterns. So by this logic:
14. Arithmetic, 15. Geometric, 16. Neither, 17. Geometric
Answer:
- <u><em>A dilation by a scale factor of 4 and then a reflection across the x-axis </em></u>
Explanation:
<u>1. Vertices of triangle FGH:</u>
- F: (-2,1)
- G: (-3,3)
- H: (0,1)
<u>2. Vertices of triangle F'G'H':</u>
- F': (-8,-4)
- G': (-12,-12)
- H': (0, -4)
<u>3. Solution:</u>
Look at the coordinates of the point H and H': to transform (0,1) to (0,-4) you can muliply each coordinate by 4 and then change the y-coordinate from 4 to -4. That is<em> a dilation by a scale factor of 4 and a reflection across the x-axis.</em> This is the proof:
- Rule for a dilation by a scale factor of 4: (x,y) → 4(x,y)
(0,1) → 4(0,1) = (0,4)
- Rule for a reflection across the x-axis:{ (x,y) → (x, -y)
(0,4) → (0,-4)
Verfiy the transformations of the other vertices with the same rule:
- Dilation by a scale factor of 4: multiply each coordinate by 4
4(-2,1) → (-8,4)
4(-3,3) → (-12,12)
- Relfection across the x-axis: keep the x-coordinate and negate the y-coordinate
(-8,4) → (-8,-4) ⇒ F'
(-12,12) → (-12,-12) ⇒ G'
Therefore, the three points follow the rules for <em>a dilation by a scale factor of 4 and then a reflection across the x-axis.</em>
