The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
Given that,
The function is (-x+3)/ (3x-2)
We have to find f(1) and f'(x).
Take the function expression
f(x)= (-x+3)/ (3x-2)
Taking x as 1 value
f(1)= (-1+3)/(3(1)-2)
f(1)=2/1
f(1)=1
Now, to get f'(x)
With regard to x, we must differentiate.
f(x) is in u/v
We know
u/v=(vu'-uv')/ v² (formula)
f'(x)= ((3x-2)(-1)- (-x+3)(3))/ (3x-2)²
f'(x)= ((-3x+2)-(-3x+9))/ 9x²- 12x+4
f'(x)=(-3x+2+3x-9)/ 9x²- 12x+4
f'(x)=2-9/ (9x²- 12x+4)
f'(x)=-7/ (9x²- 12x+4)
Therefore, The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
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Answer:
19
Step-by-step explanation:
by replacing x and y with their given values ,
we get :-
=》2x + 7y - 10
=》(2 × 4) + (7 × 3) - 10
=》8 + 21 - 10
=》29 - 10
=》19
Ohhhh nasty ! What a delightful little problem !
The first card can be any one of the 52 in the deck. For each one ...
The second card can be any one of the 39 in the other 3 suits. For each one ...
The third card can be any one of the 26 in the other 2 suits. For each one ...
The fourth card can be any one of the 13 in the last suit.
Total possible ways to draw them = (52 x 39 x 26 x 13) = 685,464 ways.
But wait ! That's not the answer yet.
Once you have the 4 cards in your hand, you can arrange them
in (4 x 3 x 2 x 1) = 24 different arrangements. That tells you that
the same hand could have been drawn in 24 different ways. So
the number of different 4-card hands is only ...
(685,464) / (24) = <em>28,561 hands</em>.
I love it !
Answer:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Step-by-step explanation:
So we have the expression:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
Recall the order of operations or PEMDAS:
P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.
E: Within the parentheses, if exponents are present, do them before all other operations.
M/D: Multiplication and division next, whichever comes first.
A/S: Addition and subtraction next, whichever comes first.
(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)
Thus, we should do the operations inside the parentheses first. Therefore:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
The parentheses is:

Square the 6 and the 4:

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

Therefore, the original equation is now:
![12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%5C%5C%3D12-%20%5B20-2%2848%29%5D)
Multiply with the brackets:
![=12-[20-96]](https://tex.z-dn.net/?f=%3D12-%5B20-96%5D)
Subtract with the brackets:
![=12-[-76]](https://tex.z-dn.net/?f=%3D12-%5B-76%5D)
Two negatives make a positive. Add:

Therefore:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Hey there Smarty!
This would be considered to be a (acute angel) which in this case, we would have to make sure that this whole triangle would equal less than 270 because each angle would be less than

°
If we add/multiply

.
So, we would have to know that was ever this would all add up to be, this would have to be less than 270°
![\left[\begin{array}{ccc}\boxed{\boxed{(2.5+5) \\ \\ (2.5*2.5) \\ (2.5-2) \\ (2.5-1) \\}} \\ \\ this \ would \ be \ why \ I \ would \ say \\ that \ this \ would \ be \ the \ answer \end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B%282.5%2B5%29%20%5C%5C%20%5C%5C%20%282.5%2A2.5%29%20%5C%5C%20%282.5-2%29%20%5C%5C%20%282.5-1%29%20%5C%5C%7D%7D%20%5C%5C%20%5C%5C%20this%20%5C%20would%20%5C%20be%20%5C%20why%20%5C%20I%20%5C%20would%20%5C%20say%20%5C%5C%20that%20%5C%20this%20%5C%20would%20%5C%20be%20%5C%20the%20%5C%20answer%20%5Cend%7Barray%7D%5Cright%5D%20%20)
I truly hope this helps, and also, it's kind of my
first time doing this I hope this would be helpful.
~Jurgen