if you look at delta y over x, you'll notice it always equals 2
i.e (22-2)/10-0 = 20/10 = 2
or (14-2)/6-0 = 12/6 = 2
or even (22-8)/10-3 = 14/7 = 2
This means that m (slope) is 2.
Now as for b. b is the y intercept and that value occurs when x = 0. On the table, when x = 0 y = 2 so b = 2.
y = mx + b becomes
y = 2x + 2
Answer:
256
Step-by-step explanation:
A calculator works well for this.
_____
None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.
1 to any power is still 1, so the factors (1^n) can be ignored.
After you ignore all of the things that can be ignored, your problem simplifies to ...
(2^2)(2^-3)^-2
The rules of exponents applicable to this are ...
(a^b)^c = a^(b·c)
(a^b)(a^c) = a^(b+c)
Then your product simplifies to ...
(2^2)(2^((-3)(-2)) = (2^2)(2^6)
= 2^(2+6)
= 2^8 = 256
Answer:
y > -x -3
Step-by-step explanation:
The graph is shaded <em>above</em> the <em>dashed</em> line, indicating y-values in the solution are greater than those on the line, so your inequality will start with ...
y >
The y-intercept of the line is (0, -3), so the "b" value in ...
y > mx +b
will be -3.
The line has a rise of -3 for a run of 3 (between the marked points), so the slope is ...
m = rise/run = -3/3 = -1
Then the inequality you want is ...
y > -x -3
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)