Answer:It's an arithmetic sequence with a common difference, d, = 4
A1 = 8
A2 = 8+4 = 12
A3 = 12+4 = 16 = 8+4(3-1) = 8+4(2) = 8+8
AN = A1 + d(N-1)
AN = 8 +4(N-1) = the Nth term in the sequence
Step-by-step explanation: hoped it helped
Answer:
0
Step-by-step explanation:
The priced was decreased by $300. So it starts at -$300. Then additional features is $900. So it's now $600. But the dealer took -$600 off. So now it is $0 from the original price.
Answer:
1. 5x +43
2. -1.5x - 7
3. 6.2 - 2x
Step-by-step explanation:
<u />
<u>Equation 1:</u>
5 (x+8) +3
First, we can distribute the 5 to the (x+8) and get 5x + 40. Distributing is when we multiply the 5 by the first number (x) and then by the second number (8) Because they aren't like terms (don't both have x's) we cannot combine then and must keep them separated by a subtraction sign
Now we have: 5x + 40 + 3
Next, we can combine the like terms. This means that any that have the same variable can be combined. So, the 5x has no other x's so he has to stay how he is. The 40 and the 3, however, can be added together to get 43.
Our finished equation is: 5x + 43
<u />
<u>Equation 2:</u>
3.6x - 7 - 5.1x
First, we can combine like terms as we learned in the last problem. This would be our x's since we have multiple.
We can add 3.6x and -5.1x and get -1.5x
Now we have: -1.5x - 7
<u />
<u>Equation 3:</u>
4 + 8x + 2.2 - 10x
We can start with either the numbers with x's or without but I'll just do the x's. So we have 8x and -10x. Adding these together would get us -2x.
Next, we can combine 4 and 2.2 and get 6.2.
Now, putting these back into our equation would look like this:
6.2 - 2x
I'm not sure how much my explanations helped, but I hope you understand!!
Answer:
2/7
Step-by-step explanation:
gradient y²-y¹/x²-x¹=4-2/4-(-3)
=2/7
{Taking (x¹,y,¹)=(-3,2)
and(x²-y²)=(4,4)
Answer:
-3/4x
Step-by-step explanation:
Down 3, right 4. We know it's negative because it's declining, and going down till you reach the x-coordinate of the point gives us -3, and going right 4 units to the location of the point.
