F(5) = - 2f(4) + 1
f(4) = -2f(3) + 1
f(3) = -2f(2) + 1
f(2) = -2f(1) + 1
Therefore:
f(2) = -2(3) + 1 = -5
f(3) = -2(-5) + 1 = 11
f(4) = -2(11) + 1 = -21
Therefore f(5) = -2(-21) + 1 = 43
If this is your equation:
![\frac{2}{5} + \frac{3}{5x} = \frac{x+5}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5x%7D%20%3D%20%5Cfrac%7Bx%2B5%7D%7B10%7D%20)
Solution:
LCD for 5 and 5x is 5x
![\frac{x}{x} ( \frac{2}{5} )+ \frac{3}{5x} = \frac{x+5}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7Bx%7D%20%28%20%5Cfrac%7B2%7D%7B5%7D%20%29%2B%20%5Cfrac%7B3%7D%7B5x%7D%20%3D%20%5Cfrac%7Bx%2B5%7D%7B10%7D%20)
![\frac{2x}{5x} + \frac{3}{5x} = \frac{x+5}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2x%7D%7B5x%7D%20%2B%20%5Cfrac%7B3%7D%7B5x%7D%20%3D%20%5Cfrac%7Bx%2B5%7D%7B10%7D%20)
![\frac{2x+3}{5x} = \frac{x+5}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2x%2B3%7D%7B5x%7D%20%3D%20%5Cfrac%7Bx%2B5%7D%7B10%7D%20)
![10(2x+3)=5x(x+5)](https://tex.z-dn.net/?f=10%282x%2B3%29%3D5x%28x%2B5%29)
←cross product
20x + 30 = 5x² + 25x ←simplify with distributive property
0 = 5x² + 5x - 30 ←use inverse operations to collect all terms on one side
0 = x² + x - 6 ←if possible divide by numerical GCF (This case 5)
0 = (x + 3)(x - 2) ←Factor
x = -3 or x = 2
Please check by substitution in original equation... Both work
Answer:
A! I am so happy to help you and hope you pass! <3
Answer: None
Step-by-step explanation: They are parallel and would never intersect
Okay I think there has been a transcription issue here because it appears to me there are two answers. However I can spot where some brackets might be missing, bear with me on that.
A direct variation, a phrase I haven't heard before, sounds a lot like a direct proportion, something I am familiar with. A direct proportion satisfies two criteria:
The gradient of the function is constant s the independent variable (x) varies
The graph passes through the origin. That is to say when x = 0, y = 0.
Looking at these graphs, two can immediately be ruled out. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations.
This leaves B and C. The graph of 1/x does not have a constant gradient, so any stretch of this graph (to y = k/x for some constant k) will similarly not be direct variation. Indeed there is a special name for this function, inverse proportion/variation. It appears both B and C are inverse proportion, however if I interpret B as y = (2/5)x instead, it is actually linear.
This leaves C as the odd one out.
I hope this helps you :)