To start, we're given the range that x lies in: from -1 to 4. We know from the fact that
that -1 will be <em /><em>included</em> in that range, so we mark -1 on the number line with a solid circle. We also know from
that, while x can be any value <em>up to</em> 4, it does not <em>include </em>4. We indicate this by drawing a hollow circle around 4 on the number line. Since x can be <em>any value within this range</em>, we make that fact clear by drawing a bold line between the points -1 and 4 on the number line. I've attached an image of what our final graph would look like.
Answer: 6x squared minus 17x minus 14
Answer:
b-4^3
Step-by-step explanation:
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:
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:
