![q(x) = {x}^{2} + 3x + 4](https://tex.z-dn.net/?f=q%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%2B%203x%20%2B%204%20)
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Step(1)
To find q(a) we just need to put a instead of x in q(x) function.
Let's do it...
![q(a) = {a}^{2} + 3a + 4](https://tex.z-dn.net/?f=q%28a%29%20%3D%20%20%7Ba%7D%5E%7B2%7D%20%2B%203a%20%2B%204%20)
Multiply sides by -2 :
![- 2q(a) = - 2( {a}^{2} + 3a + 4)](https://tex.z-dn.net/?f=%20-%202q%28a%29%20%3D%20%20-%202%28%20%7Ba%7D%5E%7B2%7D%20%2B%203a%20%2B%204%29%20)
![- 2q(a) = - 2 {a}^{2} - 6a - 8](https://tex.z-dn.net/?f=%20-%202q%28a%29%20%3D%20%20-%202%20%7Ba%7D%5E%7B2%7D%20-%206a%20-%208%20)
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Step (2)
To find q(a+1) we just need to put a+1 instead of x in q(x) function.
Let's do it...
![q(a + 1) = ({a + 1})^{2} + 3(a + 1) + 4 \\](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20%3D%20%20%28%7Ba%20%2B%201%7D%29%5E%7B2%7D%20%2B%203%28a%20%2B%201%29%20%2B%204%20%5C%5C%20%20)
![q(a + 1) = {a}^{2} + 2a + 1 + 3a + 3 + 4 \\](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20%3D%20%20%7Ba%7D%5E%7B2%7D%20%2B%202a%20%2B%201%20%2B%203a%20%2B%203%20%2B%204%20%5C%5C%20%20)
![q(a + 1) = {a}^{2} + 5a + 8](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20%3D%20%20%7Ba%7D%5E%7B2%7D%20%2B%205a%20%2B%208%20)
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Step (3)
![q(a + 1) - 2q(a) =](https://tex.z-dn.net/?f=q%28a%20%2B%201%29%20-%202q%28a%29%20%3D%20)
![{a}^{2} + 5a + 8 - 2 {a}^{2} - 6a - 8 = \\](https://tex.z-dn.net/?f=%20%7Ba%7D%5E%7B2%7D%20%2B%205a%20%2B%208%20-%202%20%7Ba%7D%5E%7B2%7D%20-%206a%20-%208%20%3D%20%20%5C%5C%20%20%20)
![- {a}^{2} - a = - a(a + 1)](https://tex.z-dn.net/?f=%20-%20%20%7Ba%7D%5E%7B2%7D%20-%20a%20%20%3D%20%20-%20a%28a%20%2B%201%29)
And we're done.
Thanks for watching buddy good luck.
♥️♥️♥️♥️♥️
The answer is 17/80
Step by step:
Move decimal 2 to the left,
.2125
Then:
2125
____
10000
Then just simplify from there
Answer:
wut?
Step-by-step explanation:
i dont know how to explain this
Answer:
from my opinion second option is right
They are both the same because anything times 0 equal 0 and 0+0 equals 0.